The maximum area of a rectangle with a perime | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the length of the rectangle be $x \ cm$ and the width be $y \ cm$.Given perimeter $P = 2(x + y) = 176 \ cm$.Therefore,$x + y = 88$,which impli

Vedclass · Accepted Answer

$1936$

Vedclass · Answer

(A) Let the length of the rectangle be $x \ cm$ and the width be $y \ cm$.Given perimeter $P = 2(x + y) = 176 \ cm$.Therefore,$x + y = 88$,which implies $y = 88 - x$.The area $A$ of the rectangle is given by $A = x 	imes y = x(88 - x) = 88x - x^2$.To find the maximum area,we differentiate $A$ with respect to $x$ and set it to zero:$\frac{dA}{dx} = 88 - 2x = 0$.Solving for $x$,we get $2x = 88$,so $x = 44 \ cm$.Since $x = 44$,then $y = 88 - 44 = 44 \ cm$.Thus,the rectangle is a square with side $44 \ cm$.The maximum area is $A = 44 	imes 44 = 1936 \ sq. \ cm$.

The maximum area of a rectangle with a perimeter of $176 \ cm$ is .......... $sq. \ cm$.

Similar Questions

The minimum distance of a point on the curve $y=x^2-4$ from the origin is

Let $f(x) = x^{2025} - x^{2000}$,$x \in [0, 1]$ and the minimum value of the function $f(x)$ in the interval $[0, 1]$ be $(80)^{80}(n)^{-81}$. Then $n$ is equal to

The minimum value of $\frac{\log x}{x}$ in the interval $[2, \infty)$ is

Slope of the tangent to the curve $y=2 e^x \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)$ where $0 \leq x \leq 2 \pi$ is minimum at $x=$

Find the local maximum and local minimum values of $f(x) = \sin 2x - x$ on the interval $[0, \pi]$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module