# An open rectangular box is to be 4 feet long and have a volume of 200 cubic feet. Find the...

## Question:

An open rectangular box is to be {eq}4 {/eq} feet long and have a volume of {eq}200 {/eq} cubic feet. Find the dimensions for which the amount of material needed to construct the box is as small as possible.

## Application of Derivative:

The above question concerns the topic of the application of derivatives to find the maxima and minima. To find the maximum or minimum value of the function, first we determine the critical point. For that, we equate the first derivative of the function to zero. The sign of the second derivative conforms to the maximum or minimum value of the function. For the positive value of the second derivative at a critical point, function attains minimum value else maximum value.

given length of the box is {eq}4 {/eq} feet long and volume of the box is {eq}200 {/eq} cubic feet.

Let the length is {eq}x {/eq} feet and width is {eq}y {/eq} feet, of the box,

The volume of the box is given as

{eq}\displaystyle \begin{align} x \times y \times 4 &= 200 \\ y& = \frac{50}{x} \end{align} {/eq}

Now the amount of material reuired to construct the box is given as

{eq}S = 2(4 \times y+ 4 \times x)+ x \times y {/eq}

Substituting the value of {eq}y {/eq}, we get

{eq}\displaystyle \begin{align} S & = 8\left ( x+ \frac{50}{x} \right ) + x \times \frac{50}{x} \\ S &= 8x+ \frac{400}{x} +50 \end{align} {/eq}

Differentiating concerning {eq}x {/eq} and equating it to zero, we get

{eq}\displaystyle \begin{align} \frac{dS}{dx} & = 8 -\frac{400}{x^2} =0 \\ x^2 &=50 \\ x&=5 \sqrt{2} \\ \end{align} {/eq}

Now {eq}\displaystyle \frac{d^2S}{dx^2} = \frac{800}{x^3}>0 {/eq} for all values of x.

So least amount of materiel used for {eq}{\color{Blue} {x= 5 \sqrt{2}}} {/eq} and {eq}{\color{Blue} {\displaystyle y =\frac{50}{5 \sqrt{2}} = 5 \sqrt{2}}} {/eq} 