Question

Determine the dimensions of the rectangle with an area 251 square feet and minimum perimeter. Round the lengths to the nearest tenth of a foot

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that,

Length of rectangle be 'x' feet

and

Breadth of rectangle be 'y' feet.

Now, it is given that

$\rm :\longmapsto\:Area_{rectangle} = 251$

$\rm :\longmapsto\:x \times y = 251$

$\bf\implies \:y = \dfrac{251}{x} - - - (1)$

Now,

Let assume that Perimeter of rectangle is P feet.

We know,

$\red{\rm :\longmapsto\:Perimeter_{rectangle} = 2(Length + Breadth)}$

So,

$\rm :\longmapsto\:P = 2(x + y)$

On substituting the value of y, we get

$\rm :\longmapsto\:P = 2 \bigg(x + \dfrac{251}{x} \bigg)$

On differentiating w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} P = 2\dfrac{d}{dx} \bigg(x + \dfrac{251}{x} \bigg)$

We know,

$\boxed{ \rm{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

$\boxed{ \rm{ \dfrac{d}{dx} \frac{1}{ {x}^{n} } = \frac{ - n}{ {x}^{n + 1} }}}$

So, using this we get

$\rm :\longmapsto\:\dfrac{dP}{dx} = 2 \bigg(1 - \dfrac{251}{ {x}^{2} } \bigg) - - - (1)$

For maxima or minima, we have

$\red{\rm :\longmapsto\:\dfrac{dP}{dx} = 0}$

$\rm :\longmapsto\: 2 \bigg(1 - \dfrac{251}{ {x}^{2} } \bigg) = 0$

$\rm :\longmapsto\: 1 - \dfrac{251}{ {x}^{2} } = 0$

$\rm :\longmapsto\: - \dfrac{251}{ {x}^{2} } = - 1$

$\rm :\longmapsto\: \dfrac{251}{ {x}^{2} } = 1$

$\rm :\longmapsto\: {x}^{2} = 251$

$\bf\implies \:x = \sqrt{251} = 15.8$

On differentiating equation (1), w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}\dfrac{dP}{dx} = 2\dfrac{d}{dx} \bigg(1 - \dfrac{251}{ {x}^{2} } \bigg)$

$\rm :\longmapsto\: \dfrac{d {}^{2} P}{d {x}^{2} } = 2 \bigg(0 + \dfrac{251 \times 2}{ {x}^{3} } \bigg)$

$\boxed{ \rm{ \because \: \dfrac{d}{dx} \frac{1}{ {x}^{n} } = \frac{ - n}{ {x}^{n + 1} } \: \: \: and \: \: \: \dfrac{d}{dx}k = 0}}$

$\rm :\longmapsto\: \dfrac{d {}^{2} P}{d {x}^{2} } = \dfrac{1004}{ {x}^{3} } > 0$

$\bf\implies \:P \: is \: minimum.$

Hence,

$\bf\implies \:y = \dfrac{251}{ \sqrt{251} } = \sqrt{251} = 15.8\: feet$

So,

$\begin{gathered}\begin{gathered}\bf\: Dimensions \: are \: \begin{cases} &\sf{Length \: = \: \sqrt{251} = 15.8 \: feet} \\ &\sf{Breadth \: = \: \sqrt{251} = 15.8\: feet} \end{cases}\end{gathered}\end{gathered}$