Chapter 4.8, Problem 53E

(a)

To determine

To write: an expression for $\mathcal{l}$ in terms of $r$ .

Answer to Problem 53E

The expression is $\mathcal{l}=\frac{1100-\pi r^2}{\pi r}$ .

Explanation of Solution

Given information:

Quonset huts are temporary, all purpose structures shaped like half cylinders and 1100 square feet of material to build a Quonset hut

The surface are $S$ of a Quonset hut is given by $S=\pi r^2+\pi r\mathcal{l}$ . Substitute 1100 for $S$ .

Calculation:

Substitute $S=1100$ in formula $S=\pi r^2+\pi r\mathcal{l}$ and then solve for $\mathcal{l}$ .

  $\begin{array}{c}\pi r^2+\pi r\mathcal{l}=1100\\ \pi r\mathcal{l}=1100-\pi r^2\\ \mathcal{l}=\frac{1100-\pi r^2}{\pi r}\end{array}$

Therefore, the expression is $\mathcal{l}=\frac{1100-\pi r^2}{\pi r}$ .

(b)

To determine

To write: an equation that gives $V$ as a function in terms of $r$ only.

Answer to Problem 53E

The equation is $V=550r-\frac{\pi }{2}{r}^3$ .

Explanation of Solution

Given information:

Quonset huts are temporary, all purpose structures shaped like half cylinders and 1100 square feet of material to build a Quonset hut

The volume $V$ of a Quonset hut is given by $V=\frac{1}{2}\pi r^2\mathcal{l}$ .

Calculation:

Substitute $\mathcal{l}=\frac{1100-\pi r^2}{\pi r}$ in formula $V=\frac{1}{2}\pi r^2\mathcal{l}$ .

  $\begin{array}{c}V=\frac{1}{2}\pi r^2\mathcal{l}\\ =\frac{1}{2}r\left(\frac{1100-\pi r^2}{\pi r}\right)\\ =\frac{1}{2}r\left(1100-\pi r^2\right)\\ =550r-\frac{\pi }{2}{r}^3\end{array}$

Therefore, the equation is $V=550r-\frac{\pi }{2}{r}^3$ .

(c)

To determine

To find: the value of $r$ that maximizes the volume of the hut.

Answer to Problem 53E

The maximum values occur when radius is about 10.8 ft.

Explanation of Solution

Given information:

Quonset huts are temporary, all purpose structures shaped like half cylinders and 1100 square feet of material to build a Quonset hut

Calculation:

Input the function using key $550X-\left(2\text{nd}\wedge X\wedge 3÷2\right)$

  

Adjust the WINDOW and hit GRAPH key.

  

Press 2^nd TRACE to go to the CALCULATE menu and highlight the Maximum and press ENTER.

  

When asked for left bound and right bound, move cursor to the left or right of observed Maximum location.

When asked for Guess, simply hit ENTER key.

  

The local maximum is at approximately $\left(10.80,3961.25\right)$ .

Therefore, the maximum values occur when radius is about 10.8 ft.