Find the largest size of a rectangle that can be inscribed in a semi circle of radius 1 unit, so that two vert

Exercise 2.4 · Q89

Q. Find the largest size of a rectangle that can be inscribed in a semi circle of radius 1 unit, so that two vertices lie on the diameter.

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Concept understanding — Maxima and Minima

Absolute (global) extrema. For $f$ defined on a domain $D$ , $f(x_0)$ is the absolute maximum of $f$ on $D$ if $f(x_0)\ge f(x)$ for every $x\in D$ ; the absolute minimum is defined symmetrically with $\le$ .

Extreme Value Theorem. If $f$ is continuous on a closed interval $[a,b]$ , then $f$ attains both an absolute maximum and an absolute minimum somewhere on $[a,b]$ — and the extremum can only occur either at an interior critical number or at one of the two endpoints.

Procedure for absolute extrema on $[a,b]$ (Exercise 7.6 Q1's method):

Find every critical number of $f$ in the open interval $(a,b)$ .
Evaluate $f$ at each critical number and at both endpoints $a,b$ .
The largest of these values is the absolute maximum; the smallest is the absolute minimum.

Relative (local) extrema. $f$ has a relative (local) maximum at $x_0$ if $f(x_0)$ is the largest value of $f$ on some open interval around $x_0$ (relative minimum: smallest, on some open interval). A function may have several local extrema, and a local extremum need not be the absolute one.

Fermat's Theorem. If $f$ has a relative extremum at $x=c$ , then $c$ must be a critical number of $f$ (so the search for local extrema always starts by solving $f'(x)=0$ together with any points where $f'$ fails to exist) — though not every critical number is automatically an extremum (e.g. $y=x^3$ at $x=0$ ).

First Derivative Test. At a critical point $c$ where $f$ is continuous, examine the sign of $f'(x)$ moving left to right across $c$ :

negative $\to$ positive: local minimum at $c$ ;
positive $\to$ negative: local maximum at $c$ ;
no sign change (same sign on both sides): $c$ is neither a local max nor a local min.

Second Derivative Test (an alternative, often quicker, at a stationary point). If $f'(c)=0$ and $f''(c)$ exists:

$f''(c)<0$ $\Rightarrow$ local maximum at $c$ ;
$f''(c)>0$ $\Rightarrow$ local minimum at $c$ ; …

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Let the rectangle have half-width $x$ (so full width $2x$ ) with its top two corners on the semicircle $x^2+y^2=1$ , so height $y=\sqrt{1-x^2}$ .

$A=2xy=2x\sqrt{1-x^2}$ .

$\dfrac{dA}{dx}=2\sqrt{1-x^2}+2x\cdot\dfrac{-x}{\sqrt{1-x^2}}=\dfrac{2(1-x^2)-2x^2}{\sqrt{1-x^2}}=\dfrac{2-4x^2}{\sqrt{1-x^2}}$ .

Setting $=0$ : $2-4x^2=0 \Rightarrow x^2=\dfrac12 \Rightarrow x=\dfrac{1}{\sqrt2}$ .

$y=\sqrt{1-\tfrac12}=\dfrac{1}{\sqrt2}$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 14 on this concept.

CBSE 2026Set V11 markMCQ
$f(x) = x$
(a) 0
(b) 1
(c) 2
(d) does not exist
›Reveal solutionSolution

On all of $\mathbb{R}$ the identity function is unbounded below, so no minimum value exists; answer (d).

For $f(x)=x$ with $x\in\mathbb{R}$ , as $x\to-\infty$ we have $f(x)\to-\infty$ . The function takes arbitrarily small (large-negative) values, so it is not bounded below an …
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CBSE 2026Set ANNUAL1 markMCQ
$x^2-y^2=4$
(a) $(3, \sqrt5)$
(b) $(2, 0)$
(c) $(\sqrt{13}, -\sqrt3)$
(d) $(\sqrt5, 1)$
›Reveal solutionSolution
Substitutes the curve's constraint into the squared-distance function and minimizes over $x$ using calculus.

A point on the curve $x^2-y^2=4$ satisfies $y^2=x^2-4$ (needs $x^2\ge4$ ).

Squared distance from $(x,y)$ to $(6,0)$ : $D^2=(x-6)^2+y^2=(x-6)^2+(x^2-4)$ .

Expand: $D^2=x^2-12x+36+x^2-4=2x^2-12x+32$ .

Differentiate w.r.t. $x$ and set to zero: $\dfrac{d(D^2)}{dx}=4x-12=0\Rightarrow x=3$ . …
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CBSE 2026Set ANNUAL1 mark
$f(x) = x \cdot \log x$
›Reveal solutionSolution

$f'(x)=\log x+1=0$ gives $x=\tfrac{1}{e}$ ; $f''>0$ confirms a minimum, and $f\!\left(\tfrac{1}{e}\right)=-\tfrac{1}{e}$ .

Differentiate $f(x) = x\log x$ using the product rule:

$f'(x) = 1\cdot\log x + x\cdot\frac{1}{x} = \log x + 1.$

Set $f'(x) = 0$ for the critical point:

$\log x + 1 = 0 \;\Rightarrow\; \log x = -1 \;\Rightarrow\; x = e^{-1} = \frac{1}{e}.$

Check the nature with the second derivative:

$f''(x) = \frac{1}{x},\qquad f''\!\left(\tfrac{1}{e}\right) = e > 0,$

…

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CBSE 2025Set X11 markMCQ
$f$
(a) $2$
(b) $0$
(c) $-2$
(d) $8$
›Reveal solutionSolution

Absolute extremum of a monotonic function on a closed interval — correct option (d).

Since $f'(x) = 3x^2 \ge 0$ , the function $f(x)=x^3$ is increasing on $[-2,2]$ , so its absolute maximum …
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CBSE 2025Set ANNUAL1 markMCQ
$x=80t-16t^2$
(a) $3$
(b) $2$
(c) $3.5$
(d) $2.5$
›Reveal solutionSolution
The stone reaches maximum height when its vertical velocity (the derivative of position) is zero; solving that gives $t=2.5$ .

Height: $x(t)=80t-16t^2$ .

Velocity: $v(t)=\dfrac{dx}{dt}=80-32t$ .

At maximum height, $v=0$ (the stone momentarily stops before falling back): $80-32t=0$ . …
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CBSE 2024Set A11 markMCQ
$f(x) = x$
(a) 1
(b) do not have maximum value
(c) 3
(d) 2
›Reveal solutionSolution

$f(x)=x$ is increasing on the open interval $(1,2)$ and its endpoint value is not attained, so there is no maximum — (b). …

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CBSE 2023Set ANNUAL1 markMCQ
$x^2e^{-2x}$
(a) $\dfrac{1}{e^2}$
(b) $\dfrac{1}{e}$
(c) $\dfrac{4}{e^4}$
(d) $\dfrac{1}{2e}$
›Reveal solutionSolution
Setting the derivative of $x^2e^{-2x}$ to zero locates the critical point $x=1$ , which gives the maximum value $1/e^2$ .

$f(x)=x^2e^{-2x}$ . By the product rule, $f'(x)=2xe^{-2x}+x^2(-2e^{-2x})=2xe^{-2x}(1-x)$ .

Setting $f'(x)=0$ for $x>0$ : since $2xe^{-2x}\ne0$ when $x>0$ , we need $1-x=0\Rightarrow x=1$ . …
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CBSE 2023Set ANNUAL1 markMCQ
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
(a) $\sqrt{ab}$
(b) $2ab$
(c) $\dfrac{a}{b}$
(d) $ab$
›Reveal solutionSolution
Parametrising the inscribed rectangle's corner on the ellipse and maximising the resulting area $2ab\sin2\theta$ gives the greatest area $2ab$ .

Let one vertex of the inscribed rectangle be $(a\cos\theta,\,b\sin\theta)$ on the ellipse, $0<\theta<\pi/2$ . By symmetry the rectangle has vertices $(\pm a\cos\theta,\pm b\sin\theta)$ . …
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CBSE 2022Set ANNUAL1 markMCQ
$|3-x|+9$
(a) $6$
(b) $0$
(c) $9$
(d) $3$
›Reveal solutionSolution
The absolute value $|3-x|$ has minimum $0$ (at $x=3$ ), so $|3-x|+9$ has minimum value $9$ .

Let $g(x)=|3-x|+9$ .

For any real $x$ , $|3-x|\ge 0$ , with the least possible value $0$ occurring exactly when $3-x=0$ , i.e. $x=3$ .

Since $9$ is a constant added to $|3-x|$ , $g(x)$ is minimized exactly when $|3-x|$ is minimized. …
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CBSE 2020Set ANNUAL1 markMCQ
$^2$
(a) $50$
(b) $10$
(c) $20$
(d) $40$
›Reveal solutionSolution
For fixed area, the perimeter of a rectangle is minimized when it is a square; with area $100$ the side is $10$ and the least perimeter is $40$ .

Let the sides of the rectangle be $x$ and $y$ , with area $xy=100$ , so $y=\dfrac{100}{x}$ .

The perimeter is $P(x)=2(x+y)=2\left(x+\dfrac{100}{x}\right)$ , for $x>0$ .

To minimize, differentiate: $P'(x)=2\left(1-\dfrac{100}{x^2}\right)$ .

Set $P'(x)=0$ : $1-\dfrac{100}{x^2}=0\Rightarrow x^2=100\Rightarrow x=10$ (taking the positive root, since $x$ is a length). …
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CBSE 2018Set ANNUAL1 markMCQ
$f$
(a) Law of mean
(b) The extreme value theorem
(c) Rolle's theorem
(d) Fermat's theorem
›Reveal solutionSolution
The stated result — a differentiable local extremum has derivative zero — is the standard statement of Fermat's theorem.

Fermat's theorem (on stationary points) states: if $f$ has a local maximum or local minimum at an interior point $c$ of its domain, and if $f'(c)$ exists, then $f'(c)=0$ .

This is exactly the statement given in the question.

Rolle's theorem is a related but different result: if $f(a)=f(b)$ and $f$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , then there exists some $c\in(a,b)$ with $f'(c)=0$ — it is about the existence of such a point between equal endpoint values, not a general statement about extrema. …
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CBSE 2017Set ANNUAL1 markMCQ
$f(x) = x^2 - 4x + 5$
(a) 2
(b) 3
(c) 4
(d) 5
›Reveal solutionSolution
On $[0,3]$ , $f(x)=x^2-4x+5$ has its minimum at the interior critical point $x=2$ and its absolute maximum at the endpoint $x=0$ , giving value $5$ .

$f'(x)=2x-4=0\Rightarrow x=2$ ; since $f''(x)=2>0$ , $x=2$ is a local minimum, with $f(2)=4-8+5=1$ .

Evaluate at the endpoints of $[0,3]$ : $f(0)=0-0+5=5$ ; $f(3)=9-12+5=2$ . …
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Q.Find the largest size of a rectangle that can be inscribed in a semi circle of radius 1 unit, so that two vertices lie on the diameter.

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