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JEECUP (UP Polytechnic) Optimization — practice questions

Explore 30 free Optimization questions in JEECUP (UP Polytechnic) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Optimization questions and exercises with answers

Showing questions 1–30 of 30.

A rectangle is inscribed in a semicircle of radius $R$ with base on the diameter. The **maximum area** is:Read the answer and worked solution →Find the **shortest distance** from origin to the curve $xy = 4$ (in the first quadrant):Read the answer and worked solution →The function $f(x) = \dfrac{x}{1 + x^2}$ attains its maximum value on $[0, \infty)$ at:Read the answer and worked solution →If $f(x) = x^x$ for $x > 0$, then $f$ has a minimum at $x = $:Read the answer and worked solution →A wire of length $L$ is bent to form a rectangle. The **maximum area** enclosed is:Read the answer and worked solution →What does the sign of f ′(x) tell you?Read the answer and worked solution →Complete the definition of a critical pointRead the answer and worked solution →Order the first derivative testRead the answer and worked solution →Match each sign change to its verdictRead the answer and worked solution →The second derivative testRead the answer and worked solution →True or false: maxima and minimaRead the answer and worked solution →Order the working rule for absolute maxima and minimaRead the answer and worked solution →Tap every point you must test for an absolute maximumRead the answer and worked solution →Local or absolute?Read the answer and worked solution →Complete the rate of change statementRead the answer and worked solution →The chain rule for related ratesRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →True or false: increasing and decreasingRead the answer and worked solution →Flip and pair: symbol with meaningRead the answer and worked solution →Match each situation to the right toolRead the answer and worked solution →Tap every statement true of a local maximum at cRead the answer and worked solution →Reading the tests backRead the answer and worked solution →Which test is being described?Read the answer and worked solution →Assertion and reason: rates of changeRead the answer and worked solution →Spot the odd oneRead the answer and worked solution →Complete the second derivative testRead the answer and worked solution →True or false: scope of this chapterRead the answer and worked solution →

Worked practice questions

A rectangle is inscribed in a semicircle of radius $R$ with base on the diameter. The **maximum area** is:

  1. $R^2$ (when the rectangle is a square inscribed)
  2. $2R^2$
  3. $R^2/2$
  4. $\pi R^2/2$
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Correct answer: A. $R^2$ (when the rectangle is a square inscribed)

Half-width = $x$, height $y = \sqrt{R^2 - x^2}$. Area $A = 2xy = 2x\sqrt{R^2 - x^2}$. We have $dA/dx = 2\sqrt{R^2 - x^2} - 2x^2/\sqrt{R^2 - x^2}$. So $dA/dx = 0$. ⇒ $R^2 = 2x^2$ ⇒ $x = R/\sqrt 2$. Then $y = R/\sqrt 2$, area = $2 \cdot (R/\sqrt 2)(R/\sqrt 2) = R^2$.

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Find the **shortest distance** from origin to the curve $xy = 4$ (in the first quadrant):

  1. 2
  2. $2\sqrt 2$
  3. 4
  4. $\sqrt 8$
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Correct answer: B. $2\sqrt 2$

Minimise $d^2 = x^2 + y^2$ subject to $xy = 4$. **By AM-GM:** $x^2 + y^2 \ge 2xy = 8$ ⇒ $d \ge \sqrt 8 = 2\sqrt 2$. Equality when $x = y = 2$ (so point is $(2, 2)$, distance $\sqrt 8 = 2\sqrt 2$).

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The function $f(x) = \dfrac{x}{1 + x^2}$ attains its maximum value on $[0, \infty)$ at:

  1. $x = 0$
  2. $x = 1$ (max value $1/2$)
  3. $x = \sqrt 3$
  4. $x \to \infty$
Show answer and solution

Correct answer: B. $x = 1$ (max value $1/2$)

We have $f'(x) = ((1+x^2) - x(2x))/(1+x^2)^2$. That is $f'(x) = (1 - x^2)/(1+x^2)^2$. So $f'(x) = 0$. ⇒ $x = 1$ (in [0,∞)). $f(1) = 1/2$. **Check:** $f(0) = 0$, $f \to 0$ as $x \to \infty$. We have **So max = 1/2 at. So $x = 1$.**.

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