Question

Find the dimensions of a rectangle with perimeter $68ft$ whose area is as large as possible.

Answer 1

Step-1:Finding the equation of dimension of a rectangle:

Let the dimension of the rectangle be length $\left(l\right)$ and width $\left(w\right)$.

We know that,

Area of the rectangle be $A=l\times w$

Perimeter of the rectangle $P=2\left(l+w\right)$

Given,

Perimeter of the rectangle $68ft$

$$\begin{gathered} \Rightarrow 68=2\left(l+w\right) \\ \Rightarrow l+w=\frac{68}{2} \\ \Rightarrow l+w=34 \\ \Rightarrow w=34-l \end{gathered}$$

Step-2:Finding the maximum area of a rectangle:

For finding the maximum value first find the critical point and then substitute in equation to find maximum value.

if $f\left(x\right)$ is a function, find $f\text{'}\left(x\right)$ and equate with zero. find $x$ value and substitute in given function

Area of the rectangle be

$$\begin{gathered} \Rightarrow A\left(l\right)=l\times \left(34-l\right) \\ =34l-l^2 \end{gathered}$$

So,

$\Rightarrow A\text{'}\left(l\right)=34-2l$

Substitute $A\text{'}\left(l\right)=0$

$$\begin{gathered} \Rightarrow 0=34-2l \\ \Rightarrow 2l=34 \\ \Rightarrow l=\frac{34}{2} \\ \Rightarrow l=17 \end{gathered}$$

$l=17$ for maximum value of area of rectangle.

Step-3:Finding the value of dimension of a rectangle:

The dimension of the rectangle be

length $l=17ft$ and

width $w=34-l\Rightarrow w=34-17=17ft$.

Hence, the dimensions of a rectangle with perimeter $68ft$ whose area is as large as possible is $l=17ft$ and $w=17ft$.