A box has a square base of side x and height y. Find the dimensions x, y for which the volume is...

Question:

A box has a square base of side {eq}x {/eq} and height {eq}y {/eq}. Find the dimensions {eq}x,\ y {/eq} for which the volume is 3 and the surface area is as small as possible.

Minima & Maxima:

The volume of the object when is fixed, then we can use the differentiation method to find the dimensions of the object that can minimize or maximize the surface area. The critical point is found first by taking the first derivative zero.

Answer and Explanation:


The volume of the box with the square base is given by:

{eq}V= x^2 y\\ {/eq}

So we are given that:

{eq}3=x^2 y\\ \Rightarrow y=\frac{3}{x^2}\\ {/eq}

Now the surface area is given as:

{eq}S=2x^2+ 4xy\\ =2x^2+ 4x \frac{3}{x^2}\\ {/eq}

Now the surface area is given as:

{eq}S=2x^2+ 4x \frac{3}{x^2}\\ =2x^2+\frac{12}{x}\\ {/eq}

Now, we will find the first derivative as zero to minimize the surface area as follows;

{eq}\frac{dS}{dx}=0\\ \Rightarrow \frac{d}{dx}\left(2x^2+\frac{12}{x}\right)=0\\ \Rightarrow 4x-\frac{12}{x^2}=0\\ \Rightarrow x=\sqrt[3]{3}\\ and ~y=\frac{3}{x^2}=\frac{3}{(\sqrt[3]{3})^2}=\sqrt[3]{3} {/eq}

So both x and y are same for minimum surface area.


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Finding Minima & Maxima: Problems & Explanation

from General Studies Math: Help & Review

Chapter 5 / Lesson 2
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