Find the dimensions of a closed rectangular box with a square base and volume 729 in^3 that can...

Question:

Find the dimensions of a closed rectangular box with a square base and volume 729 {eq}in^3 {/eq} that can be constructed with the least amount of material.

Finding Minima & Maxima:

When we are finding the maximum or the minimum value of the three dimensional shape then we have to use the calculus and its application by the differential calculus. The The critical point concept is to be also used.

Answer and Explanation:


Let the dimensions rectangular box with a square base are:

length=x

width=x

height= h

So the volume is:

{eq}x^2 h= 729\\ => h= \frac{729}{x^2}\\ {/eq}

Now the minimum material required will depend on the minimum surface area.

So now the surface area is given as;

{eq}A=2x^2+ 4xh\\ =2x^2+ 4*\frac{729}{x}\\ =2x^2+\frac{2916}{x}\\ {/eq}

Now in order to minimize this, we differentiate and get the critical point, as follows;

{eq}A'=0\\ => \frac{d}{dx}\left(2x^2+\frac{2916}{x}\right)=0\\ => 4x-\frac{2916}{x^2}=0\\ => 4x^3-2916=0\\ => x=9\\ {/eq}

So at this value the minimum surface area is obtained.

So now the dimensions of the box will be;

length=9

width=9

height= 9


Learn more about this topic:

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Finding Minima & Maxima: Problems & Explanation

from General Studies Math: Help & Review

Chapter 5 / Lesson 2
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