Chapter 12.4, Problem 4GP

To determine

Tofind:the actualtime to inflate the ball. Round to the nearest tenth.

Answer to Problem 4GP

  $8518.60inc{h}^3$ .

Explanation of Solution

Given:

Thediameterof a beachball is 18 inches.

A handpump will inflate the ball at the rate of 325 cubic inches per minute.

Calculation:

According to thequestion,

Rate of inflation of beachball is 325 cubic inches.

Now, the dimensions of cuboid and cylinder are as follows:

The dimension of cuboid is $l=30inch;w=9inch;h=17inch$ .

The dimension of cylinders is $d=25inch;h=4inch$ .

As it is known that volume of cuboid is $lwh$ and volume of the cylinder is $\pi r^{2}h$ .

Therefore, volume of cuboid is given by

  $\begin{array}{l}V=lwh\\ =\left(30\right)\left(9\right)\left(17\right)inc{h}^3\\ =4590inc{h}^3\end{array}$ .

And volume of the cylinder is given by

  $\begin{array}{l}V=\pi r^{2}h\\ =\left(\frac{22}{7}\right){\left(\frac{25}{2}\right)}^2\left(4\right)inc{h}^3\{d=25\}\\ =3928.571inc{h}^3\end{array}$ .

Now, volume of the given figure is $3928.571+4590=8518.5714inc{h}^3$

Again,volume of the given figure rounds up to nearest tenth is $8518.5714inc{h}^3=8518.60inc{h}^3$

Hence, thevolume of the given figure is $8518.60inc{h}^3$ .