Question

Your company is introducing a fruit drink packaged in an aluminum box with a square

Answer 1

Answer:

See explanation

Step-by-step explanation:

Let x in be the base side length and y in be the height of the box. Since the base is a square, we have

$S=x^2\Rightarrow x=\sqrt{S}$

The volume of the box is

$V=S\cdot y\\ \\36=Sy\Rightarrow y=\dfrac{36}{S}$

The surface area of the box is

$SA=2x^2+4xy\\ \\SA(S)=2S+4\cdot \sqrt{S}\cdot \dfrac{36}{S}=2S+\dfrac{144}{\sqrt{S}}$

The graph of the function SA(S) is shown in attached diagram.

Find the derivative of this function:

$SA'(S)=(2S+144S^{-\frac{1}{2}})'=2-\dfrac{1}{2}\cdot 144\cdot S^{-\frac{1}{2}-1}=2-\dfrac{72}{S\sqrt{S}}$

Equate this derivative to 0:

$2-\dfrac{72}{S\sqrt{S}}=0\\ \\2S\sqrt{S}=72\\ \\S\sqrt{S}=36\\ \\S^{\frac{3}{2}}=6^2\\ \\S=6^{\frac{4}{3}}$

So, the dimensions that produce a minimum surface area for this aluminum box are:

$x=\sqrt{6^{\frac{4}{3}}}=6^{\frac{2}{3}} \ in\\ \\y=\dfrac{6^2}{6^{\frac{4}{3}}}=6^{\frac{2}{3}}\ in.$