CUET UG Applied Mathematics Booster Test 2

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CUET UG Applied Mathematics Booster Test 2 - Cost, Revenue and Rates

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 2 - Cost, Revenue and Rates

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QUESTION 1 OF 20

A stationery company manufactures 'x' pens. The cost of raw material is the square of pens produced, transportation is twice the number of pens, and property tax is ₹5000. What is the pure variable cost function V(x)?

Answer Review

1 A stationery company manufactures 'x' pens. The cost of raw material is the square of pens produced, transportation is twice the number of pens, and property tax is ₹5000. What is the pure variable cost function V(x)?

A x² + 5000

B x² + 2x

C 2x + 5000

D x² + 2x + 5000

💡 Short Explanation

Variable cost changes with production. Raw material = x² and transportation = 2x. Property tax ₹5000 is fixed, so excluded.

✅ Correct Answer Explanation

Variable cost consists only of costs that vary with output. Here, raw material cost is x² and transportation cost is 2x, both depending on x. Therefore, V(x) = x² + 2x. Option B is correct. Option A includes fixed cost, C omits raw material, and D represents the total cost including fixed cost.

❓ Why Other Options Are Incorrect

Option A → x² + 5000 includes the fixed property tax and ignores transportation cost, so it is not the variable cost function.
Option C → 2x + 5000 excludes the raw material cost x² and incorrectly includes the fixed tax, making it incomplete.
Option D → x² + 2x + 5000 is the total cost function because it combines both variable and fixed costs rather than only variable costs.

🎯 Strategy

Used: Elimination

Application: Remove all options containing the fixed cost ₹5000 because variable cost excludes fixed expenses. The remaining expression correctly contains only production-dependent costs.

Final Logic: Variable cost = Raw material + Transportation = x² + 2x.

🧠 Memory Trick

"Variable = Varies with x; Fixed stays constant."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Cost Functions and Marginal Concepts

2 Given the cost data for manufacturing pens: raw material = x², transportation = 2x, tax = 5000. Match the elements in List I with List II:

List I	List I
1. Variable Cost V(x)	a. x² + 2x + 5000
2. Fixed Cost k	b. 2x + 2
3. Total Cost C(x)	c. x² + 2x
4. Marginal Cost MC	d. 5000

A 1-c, 2-d, 3-a, 4-b

B 1-a, 2-b, 3-c, 4-d

C 1-d, 2-c, 3-b, 4-a

D 1-b, 2-a, 3-d, 4-c

💡 Short Explanation

Variable cost excludes fixed cost. Total cost equals variable cost plus fixed cost. Marginal cost is the derivative of total cost.

✅ Correct Answer Explanation

Variable cost is x² + 2x (c). Fixed cost is the constant 5000 (d). Total cost is x² + 2x + 5000 (a). Differentiating total cost gives MC = dC/dx = 2x + 2 (b). Hence the correct matching is 1-c, 2-d, 3-a, 4-b, making Option A correct.

❓ Why Other Options Are Incorrect

Option B → 1-a, 2-b, 3-c, 4-d incorrectly identifies total cost as variable cost and treats marginal cost as fixed cost.
Option C → 1-d, 2-c, 3-b, 4-a mismatches every major concept. Variable cost cannot be constant, and marginal cost cannot equal total cost.
Option D → 1-b, 2-a, 3-d, 4-c incorrectly assigns marginal cost as variable cost and total cost as fixed cost, violating cost function definitions.

🎯 Strategy

Used: Option Grouping

Application: Identify each expression individually—variable cost, fixed cost, total cost, and derivative—and then match them systematically with the corresponding list.

Final Logic: Correct concepts produce the sequence 1-c, 2-d, 3-a, 4-b

🧠 Memory Trick

"V → T → MC = Variable → Total → Derivative.

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Cost Function, Total Cost and Marginal Cost

3 A firm sells 1400 units at ₹4/unit and 1800 units at ₹2/unit. Assuming a linear price-output relation \(p=mx+c\), which of the following statements are correct?
1) The slope 'm' of the price function is -1/200.
2) The linear price equation is \(p=11-\frac{1}{200}x\).
3) The revenue function \(R(x)=11x-\frac{1}{200}x^{2}\).
4) The price function is strictly positive for all \(x>2500\).

A 1, 2, 4

B 1, 2, 3

C 2, 3, 4

D 1, 3

💡 Short Explanation

Find the slope using two price-output points. Form the linear price equation. Revenue equals price × quantity; price becomes zero at x = 2200.

✅ Correct Answer Explanation

Using points (1400, 4) and (1800, 2), the slope is \((2-4)/(1800-1400)=-1/200\), so 1 is correct. Substituting gives \(p=11-\frac{x}{200}\), making 2 correct. Revenue is \(R(x)=xp=11x-\frac{x^{2}}{200}\), so 3 is correct. For \(x>2200\), price becomes negative; therefore, 4 is false because it claims the price remains positive for all \(x>2500\).

❓ Why Other Options Are Incorrect

Option A → 1, 2, 4 is incorrect because statement 4 is false. The price function becomes zero at \(x=2200\) and negative beyond that point.
Option C → 2, 3, 4 is incorrect because although 2 and 3 are true, statement 4 is false due to the negative price beyond \(x=2200\).
Option D → 1, 3 is incorrect because it omits statement 2, which correctly represents the linear price equation obtained from the given data.

🎯 Strategy

Used: Substitution

Application: Use the given coordinate pairs to determine the linear equation, then substitute into the revenue formula and verify each statement individually.

Final Logic: Since A, B, and C satisfy the calculations while D contradicts the price equation, Option B is correct.

🧠 Memory Trick

"Price × Quantity = Revenue; Revenue follows Price."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Revenue Function and Marginal Revenue

4 Identify the incorrect statement regarding price, output, and revenue combinations.

A The revenue \(R(x)\)is maximized when the first derivative \(R^{'}(x)\)equals zero and \(R^{''}(x)<0\).

B If \(p=29-x\), the revenue function is \(R(x)=29x-x^{2}\).

C The Marginal Revenue is always equal to the price 'p' regardless of the output 'x'.

D MR for \(R(x)=29x-x^{2}\)evaluates to \(29-2x\).

💡 Short Explanation

Revenue equals price × quantity. Marginal Revenue is the derivative of revenue. MR generally differs from price when price depends on output.

✅ Correct Answer Explanation

Revenue is maximized where \(R^{'}(x)=0\) and \(R^{''}(x)<0\), so A is correct. If \(p=29-x\), then \(R=x(29-x)=29x-x^{2}\), making B correct. Differentiating gives \(MR=29-2x\), confirming D. Statement C is incorrect because Marginal Revenue equals the derivative of revenue, not necessarily the price when price varies with output.

❓ Why Other Options Are Incorrect

Option A → This is the standard second derivative test for finding a maximum value and is mathematically correct.
Option B → Multiplying the price function by quantity correctly produces the revenue function \(29x-x^{2}\).
Option D → Differentiating \(29x-x^{2}\)correctly gives \(29-2x\), which is the Marginal Revenue function.

🎯 Strategy

Used: Elimination

Application: Verify each statement using derivative rules. The only statement contradicting the definition of Marginal Revenue is eliminated as incorrect.

Final Logic: Since MR equals \(dR/dx\), not always the price, Option C is the incorrect statement.

🧠 Memory Trick

"MR = dR/dx, not simply Price."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Revenue Function and Marginal Revenue

5 A firm possesses a total cost model given by \(C(x)=2x+\frac{x+4}{x+3}\). To evaluate how the cost changes dynamically, one must find the marginal cost. What rule is absolutely mandatory to find the derivative of the fractional component?

A Only the Product Rule

B Only the Chain Rule

C The Quotient Rule

D Logarithmic Differentiation only

💡 Short Explanation

Marginal cost is the derivative of total cost. The function contains a quotient of two expressions. Apply the Quotient Rule to differentiate the fractional term.

✅ Correct Answer Explanation

The total cost function is \(C(x)=2x+\frac{x+4}{x+3}\). The derivative of the linear term is straightforward, but the fractional term requires the Quotient Rule, \({\left(\frac{u}{v}\right)}^{'}=\frac{vu^{'}-uv^{'}}{v^{2}}\). Thus, Option C is correct. Option A applies to products, Option B to composite functions, and Option D is unnecessary for this algebraic fraction.

❓ Why Other Options Are Incorrect

Option A → Only the Product Rule is incorrect because the given expression is a quotient, not the product of two functions.
Option B → Only the Chain Rule is incorrect since the fraction is not primarily a composite function requiring only the Chain Rule.
Option D → Logarithmic Differentiation only is incorrect because logarithmic differentiation is generally used for complicated exponential or variable-power functions, not simple rational functions.

🎯 Strategy

Used: Elimination

Application: Identify the structure of the function first. Since it is a ratio of two algebraic expressions, eliminate rules meant for products, composites, or logarithmic forms.

Final Logic: A rational function is differentiated using the Quotient Rule, so Option C is correct.

🧠 Memory Trick

"Fraction → Quotient Rule."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Differentiation of Rational Functions and Marginal Cost

6 For the cost function \(C(x)=2x+\frac{x+4}{x+3}\), proving that the Marginal Cost (MC) falls continuously as output 'x' increases requires demonstrating that the derivative of MC (i.e., \(d^{2}C/dx^{2}\)) strictly resides in which region?

A \(d^{2}C/dx^{2}>0\)

B \(d^{2}C/dx^{2}=0\)

C \(d^{2}C/dx^{2}<0\)

D MC is constant

💡 Short Explanation

Second derivative measures the change in marginal cost. A negative second derivative means MC decreases continuously. Therefore, \(d^{2}C/dx^{2}<0\).

✅ Correct Answer Explanation

Marginal Cost is the first derivative of the total cost function. Its rate of change is given by the second derivative. If \(d^{2}C/dx^{2}<0\), Marginal Cost decreases as production increases. Hence, Option C is correct. Option A indicates increasing MC, Option B indicates no change, and Option D is inconsistent with a negative second derivative.

❓ Why Other Options Are Incorrect

Option A → \(d^{2}C/dx^{2}>0\) is incorrect because it indicates that Marginal Cost is increasing rather than decreasing.
Option B → \(d^{2}C/dx^{2}=0\) is incorrect since it means Marginal Cost remains constant, not continuously falling.
Option D → MC is constant is incorrect because a constant Marginal Cost would require the second derivative to be zero, not negative.

🎯 Strategy

Used: Conceptual/Tonal Matching

Application: Match the phrase "falls continuously" with the mathematical condition represented by the sign of the second derivative.

Final Logic: A continuously decreasing Marginal Cost requires \(d^{2}C/dx^{2}<0\), making Option C correct.

🧠 Memory Trick

"Second derivative negative → Falling slope."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Second Derivative and Rate of Change

7 If the price per unit is uniquely determined by \(p=29-x\), the revenue function formation inherently generates a curve that graphically represents a:

A Straight Line

B Downward-opening Parabola

C Upward-opening Parabola

D Cubic Curve

💡 Short Explanation

Revenue equals price × quantity. \(R(x)=x(29-x)=29x-x^{2}\). A negative coefficient of \(x^{2}\)forms a downward-opening parabola.

✅ Correct Answer Explanation

Revenue is obtained by multiplying price and quantity: \(R(x)=x(29-x)=29x-x^{2}\). Since the coefficient of \(x^{2}\)is negative, the graph is a downward-opening parabola with a maximum point. Therefore, Option B is correct. Option A is linear, Option C requires a positive quadratic coefficient, and Option D represents a cubic function.

❓ Why Other Options Are Incorrect

Option A → Straight Line is incorrect because the revenue expression contains an \(x^{2}\)term, making it quadratic rather than linear.
Option C → Upward-opening Parabola is incorrect because an upward-opening parabola requires a positive coefficient of \(x^{2}\), whereas here the coefficient is \(-1\).
Option D → Cubic Curve is incorrect since the highest power of \(x\) in the revenue function is 2, not 3.

🎯 Strategy

Used: Elimination

Application: Identify the degree and leading coefficient of the revenue function. Eliminate options representing incorrect graph types.

Final Logic: A quadratic function with a negative leading coefficient always gives a downward-opening parabola.

🧠 Memory Trick

"Negative \(x^{2}\)→ Opens Down."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Revenue Function and Maximum Revenue

8 Given \(R(x)=29x-x^{2}\), finding the continuous moving average analog in calculus (the instantaneous Marginal Revenue) gives:

A 29

B 29 - 2x

C 29x

D -2x

💡 Short Explanation

Marginal Revenue is the derivative of Revenue. Differentiate each term separately. \(R^{'}(x)=29-2x\).

✅ Correct Answer Explanation

Marginal Revenue (MR) is the first derivative of the revenue function. Differentiating \(R(x)=29x-x^{2}\)gives \(R^{'}(x)=29-2x\). Therefore, Option B is correct. Option A ignores the quadratic term, Option C is not a derivative, and Option D omits the derivative of \(29x\).

❓ Why Other Options Are Incorrect

Option A → 29 is incorrect because it differentiates only the linear term and ignores the contribution of \(-x^{2}\).
Option C → 29x is incorrect because it is not the derivative of the given revenue function and incorrectly retains the variable in the linear term.
Option D → -2x is incorrect because it omits the constant derivative obtained from differentiating \(29x\).

🎯 Strategy

Used: Substitution

Application: Apply standard differentiation rules term by term to obtain the Marginal Revenue directly from the revenue function.

Final Logic: Since \(MR=\frac{dR}{dx}=29-2x\), Option B is the correct answer.

🧠 Memory Trick

"MR = Differentiate Revenue."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Marginal Revenue

9 If the rate of change of volume of a sphere is strictly equal to the rate of change of its radius over time, what is the precise numerical value of its radius 'r'?

A 1 / (2√π)

B 2π

C 1 / (4π)

D √π / 2

💡 Short Explanation

Volume of a sphere is \(V=\frac{4}{3}\pi r^{3}\). Differentiate with respect to time. Set \(\frac{dV}{dt}=\frac{dr}{dt}\)and solve for \(r\).

✅ Correct Answer Explanation

For a sphere, \(V=\frac{4}{3}\pi r^{3}.\) Differentiating with respect to time, \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}.\) Given \(\frac{dV}{dt}=\frac{dr}{dt}\), and assuming \(\frac{dr}{dt}\neq 0\), \(4\pi r^{2}=1.\) Hence, \(r=\frac{1}{2\sqrt{\pi }}.\) Therefore, Option A is correct. Options B, C, and D do not satisfy the derived equation.

❓ Why Other Options Are Incorrect

Option B → 2π is incorrect because substituting this value into \(4\pi r^{2}=1\) does not satisfy the required condition.
Option C → 1 / (4π) is incorrect because it results from an algebraic mistake by treating \(r\) instead of \(r^{2}\).
Option D → √π / 2 is incorrect because it places \(\sqrt{\pi }\)in the numerator rather than the denominator, violating the derived relation.

🎯 Strategy

Used: Substitution

Application: Differentiate the volume formula, substitute the given condition \(\frac{dV}{dt}=\frac{dr}{dt}\), and solve the resulting equation for \(r\).

Final Logic: From \(4\pi r^{2}=1\), we obtain \(r=\frac{1}{2\sqrt{\pi }}\), making Option A correct.

🧠 Memory Trick

"Sphere → \(4\pi r^{2}\)appears after differentiation."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Related Rates of Change

10 For what values of \(x\) is the rate of increase of the total cost function
\(C(x)=x^{3}-5x^{2}+5x+8\)
exactly twice the rate of increase of \(x\)?

A \(x=2, 1/3\)

B \(x=1, 5\)

C \(x=3, 1/2\)

D \(x=1/3, 3\)

💡 Short Explanation

Differentiate the cost function. Set the derivative equal to 2. Solve the resulting quadratic equation.

✅ Correct Answer Explanation

The rate of increase of the cost function is \(C^{'}(x)=3x^{2}-10x+5.\) Since it is twice the rate of increase of \(x\), set \(C^{'}(x)=2.\) Thus, \(3x^{2}-10x+3=0=(3x-1)(x-3).\) Hence, \(x=\frac{1}{3}, 3.\) Therefore, Option D is correct. Options 1, 2, and 3 do not satisfy the equation \(C^{'}(x)=2\).

❓ Why Other Options Are Incorrect

Option A → \(x=2, 1/3\) contains the correct values but lists them in reverse order. Since the mathematical solution set is unordered, this option is also mathematically correct.
Option B → \(x=1, 5\) is incorrect because substituting either value into \(C^{'}(x)\)does not give 2.
Option C → \(x=3, 1/2\) is incorrect because \(x=\frac{1}{2}\)does not satisfy the quadratic equation obtained after differentiation.

🎯 Strategy

Used: Substitution

Application: Differentiate the function, equate the derivative to the given rate, factor the quadratic, and verify the obtained values with the options.

Final Logic: The solutions are \(x=\frac{1}{3}\)and \(x=3\). Since both A and D contain the same pair, the MCQ has duplicate correct options.

🧠 Memory Trick

"Differentiate → Equal to Rate → Factor → Solve."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Rate of Change and Marginal Cost

11 Find the exact rate of change of the lateral surface area of a cube with respect to its side \(x\), evaluated at the specific instant when \(x=4\) cm.

A 16 cm²/cm

B 32 cm²/cm

C 24 cm²/cm

D 48 cm²/cm

💡 Short Explanation

Lateral surface area of a cube is \(4x^{2}\). Differentiate with respect to \(x\). Substitute \(x=4\) to obtain the required rate.

✅ Correct Answer Explanation

The lateral surface area of a cube is \(L=4x^{2}.\) Differentiating, \(\frac{dL}{dx}=8x.\) At \(x=4\), \(\frac{dL}{dx}=8(4)=32{ cm}^{2}/cm.\) Hence, Option B is correct. Options A, C, and D do not equal the derivative evaluated at \(x=4\).

❓ Why Other Options Are Incorrect

Option A → 16 cm²/cm is incorrect because it underestimates the derivative. The correct derivative is \(8x\), which equals 32 at \(x=4\).
Option C → 24 cm²/cm is incorrect because it does not result from differentiating the lateral surface area formula correctly.
Option D → 48 cm²/cm is incorrect because it overestimates the required rate and is not obtained from \(8x\) when \(x=4\).

🎯 Strategy

Used: Substitution

Application: Differentiate the surface area formula first, then substitute the given side length into the derivative.

Final Logic: Since \(\frac{d}{dx}(4x^{2})=8x\), substituting \(x=4\) gives 32 cm²/cm.

🧠 Memory Trick

"Cube LSA = \(4x^{2}\)→ Derivative = \(8x\)."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Rate of Change in Geometry

12 A spherical iron ball 10 cm in radius is coated with ice. The ice melts at \(dV/dt=50\) cm³/min. When the ice thickness is 5 cm, the outer radius is 15 cm. What is the rate at which the thickness of ice decreases (\(dr/dt\))?

A \(1/(18\pi )\)cm/min

B \(1/(36\pi )\)cm/min

C \(1/(9\pi )\)cm/min

D \(1/(25\pi )\)cm/min

💡 Short Explanation

Ice volume depends on the outer radius. Differentiate the sphere volume formula. Use the given rate and outer radius to determine \(dr/dt\).

✅ Correct Answer Explanation

The volume of the ice layer changes according to \(V=\frac{4}{3}\pi r^{3}.\) Differentiating, \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}.\) At \(r=15\) cm, \(50=4\pi (15)^{2}\frac{dr}{dt}=900\pi \frac{dr}{dt}.\) Thus, \(\frac{dr}{dt}=\frac{50}{900\pi }=\frac{1}{18\pi } cm/min.\) Since the ice is melting, the thickness is actually decreasing, so the physical rate should be \(\frac{dr}{dt}=-\frac{1}{18\pi } cm/min.\) Therefore, the magnitude given in Option A is correct, but the sign is missing in the options.

❓ Why Other Options Are Incorrect

Option B → \(1/(36\pi )\)cm/min is incorrect because it is half of the correct magnitude obtained from differentiation.
Option C → \(1/(9\pi )\)cm/min is incorrect because it is twice the correct rate and does not satisfy the related-rate equation.
Option D → \(1/(25\pi )\)cm/min is incorrect because it is not obtained by substituting the given values into the differentiated volume formula.

🎯 Strategy

Used: Substitution

Application: Differentiate the sphere volume formula, substitute the given outer radius and melting rate, and solve for the rate of change of the radius.

Final Logic: Using

\(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt},\)
the magnitude is \(\frac{1}{18\pi }\), so Option A is numerically correct.

🧠 Memory Trick

"Sphere → Differentiate → Substitute Radius → Solve."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Related Rates of Change (Sphere)

📄 Passage

An open tank with a square bottom is to contain 4000 cubic cm of liquid. Let each side of the square base of the tank be 'x' cm and its depth be 'y' cm. Then, the volume \(V=x^{2}y=4000\), which yields \(y=4000/x^{2}\). To optimize the material used, we must minimize the surface area \(S\). The surface area consists of the square base and the four rectangular vertical sides, giving \(S=x^{2}+4xy\). By substituting \(y\) into the surface area equation, we eliminate the secondary variable, converting \(S\) into a function strictly in terms of \(x\) to apply rate of change optimization.

Finding the surface area function after substitution results in:

A \(S(x)=x^{2}+\frac{16000}{x}\)

B \(S(x)=x^{2}+\frac{4000}{x}\)

C \(S(x)=4x+\frac{16000}{x^{2}}\)

D \(S(x)=x^{3}+16000\)

💡 Short Explanation

Use the volume constraint to eliminate \(y\). Substitute \(y=\frac{4000}{x^{2}}\)into \(S=x^{2}+4xy\). Simplify to obtain a function of \(x\) only.

✅ Correct Answer Explanation

Given \(y=\frac{4000}{x^{2}},\) substitute into \(S=x^{2}+4xy.\) Thus, \(S=x^{2}+4x\left(\frac{4000}{x^{2}}\right)=x^{2}+\frac{16000}{x}.\) Hence Option A is correct. Option B omits the factor 4, Option C substitutes incorrectly, and Option D is unrelated to the given surface area formula.

❓ Why Other Options Are Incorrect

Option B → \(S(x)=x^{2}+\frac{4000}{x}\)is incorrect because the four side faces contribute \(4xy\), producing \(16000/x\), not \(4000/x\).
Option C → \(S(x)=4x+\frac{16000}{x^{2}}\)is incorrect because the substitution is performed incorrectly and does not preserve the original surface area equation.
Option D → \(S(x)=x^{3}+16000\) is incorrect because it neither follows from the substitution nor represents the tank's surface area.

🎯 Strategy

Used: Substitution

Application: Use the volume constraint to express one variable in terms of the other before simplifying the surface area expression.

Final Logic: Replacing \(y\) with \(\frac{4000}{x^{2}}\)directly gives

\(S(x)=x^{2}+\frac{16000}{x}.\)

🧠 Memory Trick

"Volume first, substitute next, optimize last."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Optimization Problems (Open Tank)

📄 Passage

To achieve the minimum surface area, taking the derivative
\(S^{'}(x)=2x-\frac{16000}{x^{2}}\)
and equating to zero provides the optimal base dimension 'x' as:

A 10 cm

B 20 cm

C 30 cm

D 40 cm

💡 Short Explanation

Differentiate the surface area function. Set the derivative equal to zero. Solve the resulting cubic equation for \(x\).

✅ Correct Answer Explanation

From \(S(x)=x^{2}+\frac{16000}{x},\) we obtain \(S^{'}(x)=2x-\frac{16000}{x^{2}}.\) Setting \(S^{'}(x)=0\), \(2x=\frac{16000}{x^{2}}2x^{3}=16000x^{3}=8000x=20 cm.\) Therefore, Option B is correct. The remaining options do not satisfy the optimization equation.

❓ Why Other Options Are Incorrect

Option A → 10 cm is incorrect because substituting \(x=10\) does not make the first derivative zero.
Option C → 30 cm is incorrect since \({30}^{3}\neq 8000\), so it cannot satisfy the critical-point equation.
Option D → 40 cm is incorrect because it produces a much larger value than required and does not minimize the surface area.

🎯 Strategy

Used: Substitution

Application: Differentiate the surface area function, equate the derivative to zero, and solve the resulting equation systematically.

Final Logic: Since

\(x^{3}=8000,\)
the optimal base side is 20 cm, making Option B correct.

🧠 Memory Trick

"Differentiate → Equal zero → Cube root."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Optimization of Surface Area

15 In shadow and ladder geometry, if a 10 m ladder slips such that \(x^{2}+y^{2}=100\), and the lower end \(x\) moves away at \(\frac{dx}{dt}=2\) m/min, what is the rate \(\frac{dy}{dt}\)at which the upper end falls exactly when \(x=6\) m (and consequently \(y=8\) m)?

A -2 m/min

B -3/4 m/min

C -3/2 m/min

D -4/3 m/min

💡 Short Explanation

Differentiate the ladder equation implicitly. Substitute the given values of \(x\), \(y\), and \(\frac{dx}{dt}\). Solve for \(\frac{dy}{dt}\).

✅ Correct Answer Explanation

The ladder satisfies \(x^{2}+y^{2}=100.\) Differentiating with respect to time, \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0.\) Hence, \(\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}.\) Substituting \(x=6\), \(y=8\), and \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=-\frac{6}{8}\times 2=-\frac{12}{8}=-\frac{3}{2} m/min.\) Therefore, Option C is correct. The negative sign indicates that the upper end is moving downward.

❓ Why Other Options Are Incorrect

Option A → -2 m/min is incorrect because it ignores the ratio \(\frac{x}{y}\)that appears after implicit differentiation.
Option B → -3/4 m/min is incorrect because it omits multiplication by the given rate \(\frac{dx}{dt}=2\).
Option D → -4/3 m/min is incorrect because it incorrectly uses the reciprocal ratio \(\frac{y}{x}\).

🎯 Strategy

Used: Substitution

Application: Differentiate the constraint equation implicitly and substitute the numerical values directly into the derived formula.

Final Logic: Using

\(\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt},\)
gives \(-\frac{3}{2}\)m/min, so Option C is correct.

🧠 Memory Trick

"Differentiate → Substitute → Negative means falling."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Related Rates (Ladder Problem)

16 A boy 1 m tall walks away from a 5 m lamp post. If the ratio of shadow \(y\) to total distance \(x+y\) is constrained by similar triangles as
\(\frac{y}{x+y}=\frac{1}{5},\)
this strictly simplifies to the linear dependent relation:

A \(y=4x\)

B \(y=\frac{x}{4}\)

C \(y=5x\)

D \(y=\frac{x}{5}\)

💡 Short Explanation

Apply the property of similar triangles. Cross-multiply the given proportion. Rearrange to obtain the relation between \(x\) and \(y\).

✅ Correct Answer Explanation

Given \(\frac{y}{x+y}=\frac{1}{5},\) cross-multiplying gives \(5y=x+y.\) Rearranging, \(4y=x.\) Hence, \(y=\frac{x}{4}.\) Therefore, Option B is correct. Options A, C, and D do not satisfy the proportion derived from similar triangles.

❓ Why Other Options Are Incorrect

Option A → \(y=4x\) is incorrect because it reverses the relationship obtained after rearranging the equation.
Option C → \(y=5x\) is incorrect because it ignores the subtraction of \(y\) from both sides after cross-multiplication.
Option D → \(y=\frac{x}{5}\)is incorrect because it incorrectly assumes the denominator remains unchanged during simplification.

🎯 Strategy

Used: Substitution

Application: Cross-multiply the proportion and simplify algebraically to express one variable in terms of the other.

Final Logic: From

\(5y=x+y,\)
we obtain
\(4y=x,\)
so Option B is correct.

🧠 Memory Trick

"Similar triangles → Cross multiply → Rearrange."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Related Rates (Shadow Problems)

17 For the cubic cost model
\(C(x)=300x-10x^{2}+\frac{1}{3}x^{3},\)
finding the minimum Marginal Cost requires deriving MC to get \(MC^{'}(x)=-20+2x\). This indicates the Marginal Cost reaches its absolute floor at an output of:

A \(x=20\)

B \(x=10\)

C \(x=30\)

D \(x=5\)

💡 Short Explanation

Marginal Cost is the first derivative of cost. Minimum MC occurs where \(MC^{'}(x)=0\). Solve the resulting linear equation for \(x\).

✅ Correct Answer Explanation

Differentiate the cost function: \(MC=C^{'}(x)=300-20x+x^{2}.\) Differentiate again: \(MC^{'}(x)=2x-20.\) For the minimum Marginal Cost, \(MC^{'}(x)=0\) gives \(2x-20=0\Rightarrow x=10.\) Also, \(MC^{''}(x)=2>0,\) confirming a minimum. Therefore, Option B is correct. The remaining options do not satisfy the critical-point condition.

❓ Why Other Options Are Incorrect

Option A → \(x=20\) is incorrect because \(MC^{'}(20)=20\neq 0\), so it is not a stationary point.
Option C → \(x=30\) is incorrect since it does not satisfy the equation \(2x-20=0\).
Option D → \(x=5\) is incorrect because \(MC^{'}(5)=-10\neq 0\), indicating Marginal Cost is still decreasing.

🎯 Strategy

Used: Substitution

Application: Differentiate twice, equate the first derivative of Marginal Cost to zero, and verify the minimum using the second derivative.

Final Logic: Since

\(MC^{'}(x)=2x-20=0,\)
the minimum Marginal Cost occurs at \(x=10\).

🧠 Memory Trick

"Minimum → First derivative zero, second derivative positive."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Marginal Cost and Optimization

18 A tour operator charges ₹136 per passenger with a discount. If the structured revenue model shifts based on passenger constraints, calculating the Marginal Revenue dynamically requires utilizing the:

A Constant Rule exclusively

B Product Rule exclusively

C First Derivative Test

D Integration Rules

💡 Short Explanation

Marginal Revenue equals the derivative of Revenue. Differentiate the revenue function with respect to output. The First Derivative Test is used for extrema, not for computing MR.

✅ Correct Answer Explanation

Marginal Revenue (MR) is defined as \(MR=\frac{dR}{dx},\) the first derivative of the revenue function. The First Derivative Test is a technique for identifying maxima and minima after differentiation, not for calculating MR itself. Therefore, Option C is conceptually incorrect. Options A, B, and D are also incorrect because MR is obtained by ordinary differentiation, not by constant rule alone, product rule alone, or integration.

❓ Why Other Options Are Incorrect

Option A → Constant Rule exclusively is incorrect because revenue functions generally contain variable terms, not only constants.
Option B → Product Rule exclusively is incorrect because not every revenue function requires the Product Rule. Many require simple differentiation or expansion first.
Option D → Integration Rules is incorrect because integration is the inverse of differentiation and is not used to calculate Marginal Revenue.

🎯 Strategy

Used: Elimination

Application: Recall the definition of Marginal Revenue and eliminate options that do not describe differentiation correctly.

Final Logic: Marginal Revenue is found by differentiating the revenue function. Since no option states this correctly, the MCQ contains a conceptual error.

🧠 Memory Trick

"MR = dR/dx."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Marginal Revenue

19 A manufacturer produces \(x\) pants with
\(C(x)=x^{2}+78x+2500\)
and price
\(p=600-8x.\)
The comprehensive Profit function \(P(x)=R(x)-C(x)\)evaluates to a quadratic. At what output \(x\) is the profit maximized (where \(P^{'}(x)=0\))?

A 25

B 29

C 35

D 40

💡 Short Explanation

Revenue equals price × quantity. Profit equals Revenue − Cost. Differentiate the profit function and equate to zero.

✅ Correct Answer Explanation

Revenue is \(R(x)=x(600-8x)=600x-8x^{2}.\) Thus, \(P(x)=R(x)-C(x)=-9x^{2}+522x-2500.\) Differentiating, \(P^{'}(x)=-18x+522.\) Setting \(P^{'}(x)=0\), \(x=\frac{522}{18}=29.\) Since \(P^{''}(x)=-18<0\), the profit is maximum at \(x=29\). Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → 25 is incorrect because substituting \(x=25\) does not satisfy the condition \(P^{'}(x)=0\).
Option C → 35 is incorrect because the first derivative is not zero at this value, so it cannot maximize profit.
Option D → 40 is incorrect because the profit function is already decreasing beyond the optimal output of 29.

🎯 Strategy

Used: Substitution

Application: Form the profit function, differentiate it, and solve the resulting linear equation to locate the critical point.

Final Logic: Since

\(P^{'}(x)=-18x+522=0,\)
the maximum profit occurs at \(x=29\).

🧠 Memory Trick

"Profit = Revenue − Cost → Differentiate → Set to Zero."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Profit Maximization

20 Continuing with the profit function
\(P(x)=-9x^{2}+522x-2500,\)
substituting the optimal output \(x=29\) yields the absolute maximum weekly profit of:

A ₹5069

B ₹2500

C ₹7569

D ₹4962

💡 Short Explanation

Substitute the optimal output into the profit function. Perform arithmetic carefully. The resulting value is the maximum profit.

✅ Correct Answer Explanation

Using \(P(x)=-9x^{2}+522x-2500,\) at \(x=29\), \(P(29)=-9(29)^{2}+522(29)-2500.\) Since \({29}^{2}=841,-9(841)=-7569,522(29)=15138,\) we get \(P(29)=15138-7569-2500=5069.\) Therefore, Option A is correct. The remaining options are obtained from incorrect arithmetic.

❓ Why Other Options Are Incorrect

Option B → ₹2500 is incorrect because it represents only the constant term in the profit function, not the evaluated maximum profit.
Option C → ₹7569 is incorrect because it equals \(9\times {29}^{2}\), ignoring the remaining terms in the expression.
Option D → ₹4962 is incorrect because it results from an arithmetic error while evaluating the quadratic function.

🎯 Strategy

Used: Substitution

Application: Substitute the optimal value of \(x\) obtained in the previous question into the profit function and simplify accurately.

Final Logic: Evaluating

\(P(29)\)
gives ₹5069, so Option A is correct.

🧠 Memory Trick

"Find optimum first, substitute next."

📚 Class XII – Mathematics Part II – Chapter: Applications of Derivatives – Topic: Maximum Profit

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