Chapter 7.5, Problem 23E

To determine

To check: if hiring the firm will be affordable.

Answer to Problem 23E

Yes, that can be afforded.

Explanation of Solution

Given information:

The figure is:

  

The hemispherical tank has radius $10ft$ and the Benzene weights $56{\text{ft/lb}}^{\text{3}}$ .

Formula used:

The work to move an object from $x=a$ to $x=b$ is: $W={\displaystyle \underset{a}{\overset{b}{\int }}F(x)dx}$ .

  $F\left(x\right)$ is weight of water above the pump, that is, $F\left(x\right)=\gamma x$ .

  $1\text{cent}=0.01\text{dollars}$ .

Calculation:

The work to raise the thin disk is

  $\pi r^2(56)h=56\pi {(\sqrt{{10}^2-y^2})}^2(12-y)\Delta y$ .

Therefore, $F(x)=56\pi (12-y)(100-y^2);a=0;b=10$

The total work is:

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{10}{\int }}56\pi \left(12-y\right)\left(100-y^2\right)}dy=56\pi {\displaystyle \underset{0}{\overset{10}{\int }}\left(1200-12y^2-100y+y^3\right)}dy\\ =56\pi {\left[1200y-4y^3-50y^2+\frac{y^4}{4}\right]}_0^{10}\\ =56\pi \left[1200\left(10\right)-4{\left(10\right)}^3-50{\left(10\right)}^2+\frac{{\left(10\right)}^4}{4}-0\right]\\ =56\pi \left(5500\right)\\ =967611.5373\\ \approx 967611\text{ft-lb}\end{array}$

Now, $0.5$ cent = $0.005$ dollars

Using the Cent to dollar conversion, the total amount required as,

  $\$(967611)(0.005)\approx \$4838$ .

As it is less than $\$5000$ , hence it can be afforded.