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📝 Applied Maximum and Minimum Problems in calculus (24 MCQs)

📖 From Calculus • 5. The derivative in Graphing and Applications • 24 questions available

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What is Applied Maximum and Minimum Problems in calculus?

Definition:
Applied optimization involves modeling real-world scenarios with functions, then finding absolute extrema to maximize profit, minimize cost, or optimize dimensions. Constraints define the domain, and derivatives identify critical points that yield optimal solutions for practical engineering or economic problems.

Example:
To maximize area of a rectangle with perimeter $20$ , let $A = x(10-x)$ . $A' = 10-2x=0 \Rightarrow x=5$ , giving square $5\times5$ with max area $25$ .

Reason:
Translating physical constraints into mathematical functions allows calculus to find efficient solutions, bridging theoretical math with tangible real-world applications.

Easy

Medium

Hard

📝 All Applied Maximum and Minimum Problems in calculus MCQs

Q1. If the perimeter of a rectangular garden is fixed at 100 ft, which of the following statements about the area as a function of length $x$ is true?

A.Area increases linearly with

x

B.Area is a quadratic function opening upward

C.Area is a quadratic function opening downward ✅

D.Area is constant

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The area expression $A(x)=x(50-x)=50x-x^{2}$ is a quadratic polynomial whose leading coefficient is negative, so the parabola opens downward. Therefore the correct description is that the area is a quadratic function opening downward.

Q2. Given the relation $y = 50 - x$ for the garden dimensions, what happens to the width $y$ when the length $x$ increases from 10 ft to 20 ft?

y

increases by 10 ft

y

decreases by 10 ft ✅

y

remains unchanged

y

becomes negative

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Substituting $x=10$ gives $y=40$ ; substituting $x=20$ gives $y=30$ . The width drops by $10$ ft, showing a direct inverse linear relationship between length and width.

Q3. If a rectangle with perimeter 100 ft has maximum area, what must be true about its sides?

A.Length equals width ✅

B.Length is twice width

C.Length is half width

D.Length is zero

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The maximum area occurs when the rectangle is a square. Setting the derivative of $A(x)=x(50-x)$ to zero yields $x=25$ ft, which equals the width, confirming that length equals width.

Q4. Suppose a rectangular garden uses 100 ft of fence and the length is 30 ft. What is the area compared to the maximum possible area?

A.It is larger

B.It is equal

C.It is smaller ✅

D.Cannot be determined

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: With length 30 ft, width is $y=50-30=20$ ft, giving area $30\times20=600$ ft². The maximum possible area is $25\times25=625$ ft², so the given area is smaller.

Q5. If the length $x$ is constrained to be an integer, which integer value of $x$ in $[0,50]$ yields the greatest area?

A.24

B.25 ✅

C.26

D.27

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The area function $A(x)=x(50-x)$ attains its maximum at $x=25$ . Since 25 is an integer within the interval, it provides the greatest integer‑based area of $625$ ft².

Q6. Assume a mistake in measuring fence results in 2 ft extra. How does the maximum possible area change?

A.Increases by 25.25 ft² ✅

B.Increases by 12.5 ft²

C.Increases by 5 ft²

D.No change

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: With 102 ft of fence, the optimal side length becomes $102/4=25.5$ ft, giving area $25.5^{2}=650.25$ ft². The original maximum was $625$ ft², so the increase is $650.25-625=25.25$ ft².

Q7. Compare the area function $A(x)=x(50-x)$ with the function $B(x)=x^{2}$ . Which statement is true for $0<x<25$ ?

A(x) > B(x)

✅

A(x) = B(x)

A(x) < B(x)

D.Cannot compare

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For $0<x<25$ , $A(x)=50x-x^{2}$ exceeds $x^{2}$ because the linear term $50x$ dominates the quadratic term, making $A(x)$ larger than $B(x)=x^{2}$ .

Q8. Which of the following correctly describes the vertex of the parabola representing the area function?

A.At

(0,0)

B.At

(25,625)

✅

C.At

(50,0)

D.At

(25,0)

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Writing $A(x)= -x^{2}+50x$ in vertex form gives $-\bigl(x-25\bigr)^{2}+625$ . Hence the vertex is at $(25,625)$ , the point of maximum area.

Q9. Evaluate the effect on maximum area if the perimeter is reduced by 20%.

A.Area reduces by 20%

B.Area reduces by 36% ✅

C.Area reduces by 40%

D.Area reduces by 64%

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Original maximum area is $(100/4)^{2}=625$ ft². Reducing perimeter to 80 ft gives $(80/4)^{2}=400$ ft². The reduction is $(625-400)/625=0.36$ , i.e., a 36% decrease.

Q10. If the rectangle is changed to a square with side $s$ , and the same fence is used, what is the relationship between $s$ and the original length $x$ that gives the same area?

s = x

s = 50 - x

s = \sqrt{x(50-x)}

✅

s = \dfrac{x+50}{2}

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Equating the square’s area $s^{2}$ to the rectangle’s area $x(50-x)$ yields $s^{2}=x(50-x)$ . Solving for $s$ gives the positive root $s=\sqrt{x(50-x)}$ .

Q11. Given the area function $A(x)=50x - x^{2}$ , determine the second derivative and its implication for the nature of the critical point.

A&#039;&#039; = -2

, indicating a maximum ✅

A&#039;&#039; = 2

, indicating a minimum

A&#039;&#039; = 0

, inconclusive

A&#039;&#039; = -50

, indicating a maximum

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Differentiating once gives $A' = 50-2x$ ; differentiating again yields $A'' = -2$ , a constant negative value, confirming the critical point is a maximum.

Q12. Suppose the garden must have length at least 20 ft. Under this constraint, what is the maximum possible area?

A.600 ft²

B.625 ft² ✅

C.560 ft²

D.500 ft²

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The unrestricted maximum occurs at $x=25$ ft, which satisfies the $x\ge20$ condition. Hence the maximum area remains $25\times25 = 625$ ft².

Q13. Applying the principle that among all rectangles with a given perimeter the square has the largest area, which shape does the garden become at maximum area?

A.Rectangle with length twice width

B.Square ✅

C.Circle

D.Triangle

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The theorem states that a square encloses the greatest area for a fixed perimeter. Therefore the garden must be a square when the area is maximized.

Q14. Which calculus concept is used to find the maximum area of the garden?

A.Integration

B.Limits

C.Derivative ✅

D.Series

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: To locate the maximum of the area function $A(x)=x(50-x)$ , we set its derivative $A'(x)$ to zero. This use of the derivative identifies stationary points.

Q15. If a farmer wants to enclose a rectangular field using 200 ft of fencing, how does the maximum possible area compare to that of the 100 ft case?

A.Four times larger ✅

B.Eight times larger

C.Two times larger

D.Same size

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Maximum area scales with the square of the perimeter: $(200/4)^{2}=2500$ ft² versus $(100/4)^{2}=625$ ft², giving a factor of $2500/625=4$ .

Q16. Explain why the derivative set to zero gives the maximum area for this problem.

A.It identifies where slope is zero, indicating a horizontal tangent, which for a concave‑down parabola is a maximum. ✅

B.It finds where the function is undefined.

C.It finds where the function equals zero.

D.It gives the minimum area.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When the derivative equals zero, the function’s graph has a horizontal tangent. Because the area function is a downward‑opening parabola, this stationary point must be the vertex, which is the global maximum.

Q17. If the fence material costs \$2 per foot, how does the cost change when the garden is built to achieve maximum area versus any other dimensions?

A.Cost is higher at maximum area

B.Cost is lower at maximum area

C.Cost is the same because perimeter is fixed ✅

D.Cost depends on shape

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The total length of fence is fixed at 100 ft, so regardless of the rectangle’s dimensions the material cost remains $100 \times \<span class="katex-error" title="ParseError: KaTeX parse error: Unexpected character: '\' at position 5: 2 = \̲" style="color:#cc0000">2 = \</span>200$ . Hence cost is unchanged.

Q18. A rectangular garden is to be built with a fixed perimeter, but the length must be 10% longer than the width. What is the resulting maximum area relative to the unrestricted maximum?

A.99.9% of unrestricted max ✅

B.95% of unrestricted max

C.90% of unrestricted max

D.80% of unrestricted max

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Let width $w$ ; length $1.1w$ . Perimeter $2(w+1.1w)=4.2w=100$ gives $w\approx23.81$ ft, area $1.1w^{2}\approx624.6$ ft². Unrestricted maximum is 625 ft², yielding about 99.9% of the full maximum.

Q19. Consider the function $A(x)=x(50-x)$ . If we introduce a penalty term $-0.5x^{2}$ to model diminishing returns, the new function is $A'(x)=x(50-x)-0.5x^{2}$ . What $x$ maximizes $A'$ ?

A.16.67 ✅

B.20

C.25

D.30

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Simplifying gives $A'(x)=50x-1.5x^{2}$ . Differentiating yields $A''(x)=50-3x$ . Setting this to zero gives $x=50/3\approx16.67$ ft, the point of maximum for the modified function.

Q20. What is the formula for the perimeter of a rectangle with length $x$ and width $y$ ?

P = 2x + y

P = x + 2y

P = 2x + 2y

✅

P = xy

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The perimeter of a rectangle adds the lengths of all four sides: two lengths $x$ and two widths $y$ , giving $P = 2x + 2y$ .

Q21. Who is credited with developing the first procedure for differentiating polynomials?

A.Isaac Newton

B.Gottfried Leibniz

C.Pierre de Fermat ✅

D.René Descartes

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Pierre de Fermat introduced a systematic method for finding rates of change of polynomial expressions, effectively creating the first differentiation technique.

Q22. In the example, what is the value of $x$ at which the derivative $\dfrac{dA}{dx}$ equals zero?

A.10 ft

B.25 ft ✅

C.30 ft

D.50 ft

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: From $A(x)=50x-x^{2}$ , the derivative is $\dfrac{dA}{dx}=50-2x$ . Setting it to zero gives $x=25$ ft, the point where the area is maximized.

Q23. If the garden's length is set to 15 ft, what percent of the maximum area is achieved?

A.0.84 ✅

B.0.8

C.0.7

D.0.6

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Area at $x=15$ ft is $15\times35=525$ ft². The maximum area is 625 ft². The percentage is $\frac{525}{625}\times100=84\%$ .

Q24. If the fence is used to enclose a rectangle and a semicircle attached to one side, how does the maximum possible rectangular area compare to the pure rectangle case?

A.It is larger

B.It is the same

C.It is smaller ✅

D.Cannot be determined without more info

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Adding a semicircle consumes part of the fixed perimeter, leaving less fence for the rectangle. Consequently, the rectangle’s maximum achievable area must be smaller than when the entire perimeter is devoted to a rectangle alone.

🔗 Related Topics (MCQs)

📝 Increasing and decreasing functions using derivative MCQs 📝 Concavity and inflection points calculus MCQs 📝 Logistic growth curves calculus MCQs 📝 Analysis of Functions II (Relative Extrema) in calculus MCQs 📝 Relative maxima and minima local extrema MCQs

←Previous Topic:Absolute extrema with one critical point📝 Applied Maximum and Minimum Problems in calculusOptimization problems on closed intervals:Next Topic→