A wire of length 64 text{ m} is to be bent to | vedclass

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

Question

(C) Let the length and breadth of the rectangle be $l$ and $b$ respectively.The perimeter of the rectangle is $2(l + b) = 64 	ext{ m}$, so $l + b = 3

Vedclass · Accepted Answer

$256$

Vedclass · Answer

(C) Let the length and breadth of the rectangle be $l$ and $b$ respectively.The perimeter of the rectangle is $2(l + b) = 64 	ext{ m}$, so $l + b = 32 	ext{ m}$, which implies $b = 32 - l$.The area $A$ is given by $A = l 	imes b = l(32 - l) = 32l - l^2$.To find the maximum area, we differentiate $A$ with respect to $l$: $\frac{dA}{dl} = 32 - 2l$.Setting $\frac{dA}{dl} = 0$, we get $32 - 2l = 0$, so $l = 16 	ext{ m}$.Then $b = 32 - 16 = 16 	ext{ m}$.The maximum area is $A = 16 	imes 16 = 256 	ext{ m}^2$.

$A$ wire of length $64 \text{ m}$ is to be bent to form a rectangle such that its area is maximum. What is the maximum area (in $\text{ m}^2$)?

Similar Questions

The volume of the largest possible right circular cylinder that can be inscribed in a sphere of radius $R = \sqrt{3}$ is

If $y = a \log x + bx^2 + x$ has extreme values at $x = 1$ and $x = 2$,then $(a, b) = \dots$

The sum of the absolute maximum and minimum values of the function $f(x) = |x^2 - 5x + 6| - 3x + 2$ in the interval $[-1, 3]$ is equal to:

The function $f(x) = x e^{-x}$ for all $x \in R$ attains a maximum value at $x = k$, then $k = $

If $f(x) = 2x^3 - 21x^2 + 36x - 30$,then which one of the following is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module