Chapter 5.3, Problem 32PS

Solution

(a)

To write an equation that gives height (h) in terms of length (s).

  $\text{h=}\frac{16\sqrt{3}}{{\text{3s}}^{\text{2}}}$

Given information: $\text{V=}\frac{\text{3}\sqrt{\text{3}}}{\text{2}}{\text{s}}^{\text{2}}\text{h}$

Formula used: Rationalization

Calculation:

Solve for  $\text{h}$  to get the equation in terms of  $\text{s}$.

Multiplying by $\frac{\sqrt{\text{3}}}{\sqrt{\text{3}}}$ to get rid of the square root in the denominator.

  $\begin{array}{c}\text{V=24=}\frac{\text{3}\sqrt{\text{3}}}{\text{2}}{\text{s}}^{\text{2}}\text{h}\\ \text{24}\text{.}\frac{\text{2}}{\text{3}\sqrt{\text{3}}{\text{s}}^{\text{2}}}\text{=h}\\ \text{h=}\frac{\text{16}}{\sqrt{\text{3}}{\text{s}}^{\text{2}}}\text{.}\frac{\sqrt{\text{3}}}{\sqrt{\text{3}}}\\ \text{=}\frac{\text{16}\sqrt{\text{3}}}{{\text{3s}}^{\text{2}}}\end{array}$

Hence, $\text{h=}\frac{\text{16}\sqrt{\text{3}}}{{\text{3s}}^{\text{2}}}$

(b)

Find the dimensions length (s) and height (h) that minimize the amount of material used

Minimum volume is $37.77$

From the hint with part(a), we know that

  $\text{V=}\frac{\text{3}\sqrt{\text{3}}}{\text{2}}{\text{s}}^{\text{2}}\text{h}$

But the volume is given as 24

Therefore

  $\begin{array}{c}\text{24=}\frac{\text{3}\sqrt{\text{3}}}{\text{2}}{\text{s}}^{\text{2}}\text{h}\\ \text{h=}\frac{\text{16}}{\sqrt{\text{3}}{\text{s}}^{\text{2}}}\end{array}$

In Hint with sub-part(b) it is given that

  $\text{S=}\frac{\text{3}\sqrt{\text{3}}}{\text{2}}{\text{s}}^{\text{2}}\text{+6sh}$

Substitute the value of  $\text{h}$

  $\begin{array}{c}\text{S=}\frac{\text{3}\sqrt{\text{3}}}{\text{2}}{\text{s}}^{\text{2}}+6s\left(\frac{16}{\sqrt{3}{s}^2}\right)\\ \text{S=}\frac{\text{3}\sqrt{\text{3}}}{\text{2}}{\text{s}}^{\text{2}}+\frac{96}{\sqrt{3s}}\end{array}$

Graph is shown below:

  

From the above graph we can see that the minimum volume is $37.77$

Minimum volume is $37.77$