Answer to Question #285817 in Calculus for john

Question

SHOW COMPLETE SOLUTION

A closed cylindrical tank has a capacity of 50.265 m^3.

A. determine the radius of the cylindrical tank that requires minimum amount of material used.

B. compute the height of the tank.

C. if the galon of paint covers an area of 28 m^2, compute the number of gallons needed to paint the tank using two coats of paint.

Answer 1

A. Let $r=$ the radius of the base of the cylinder, let $h=$ the height of the cylinder. Then $V=\pi r^2 h.$ Solve for $h$

$$h=\dfrac{V}{\pi r^2}$$

The amount of the material is

$$A=2\pi r^2+2\pi r h$$

Substitute

$$A=2\pi r^2+\dfrac{2V}{ r}$$

Given $V=50.265 m^3.$ Then

$$A=A(r)=2\pi r^2+\dfrac{2(50.265)}{ r}, m, r>0$$

Find the first derivative with respect to $r$

$$A'(r)=4\pi r-\dfrac{100.53}{ r^2}$$

Find the critical number(s)

$$A'=0=>4\pi r-\dfrac{100.53}{ r^2}=0$$

$$r=\sqrt[3]{\dfrac{100.53}{ 4\pi}}\approx2.000\ m$$

If $0<r<\sqrt[3]{\dfrac{100.53}{ 4\pi}}, A'(r)<0, A(r)$ decreases.

If $r>\sqrt[3]{\dfrac{100.53}{ 4\pi}}, A'(r)>0, A(r)$ increases.

The function $A(r)$ has a local minimum at $r=\sqrt[3]{\dfrac{100.53}{ 4\pi}}.$

Since the function $A(r)$ has the only extremum, then the function $A(r)$ has the absolute minimum at $r=\sqrt[3]{\dfrac{100.53}{ 4\pi}}.$

The radius of the cylindrical tank that requires minimum amount of material used is $2$ m.

B.

$$h=\dfrac{V}{\pi r^2}=\dfrac{Vr}{\pi r^3}$$

If $r=\sqrt[3]{\dfrac{100.53}{ 4\pi}},$  then

$$h=\dfrac{V}{\pi r^2}=\dfrac{50.265}{\pi (\dfrac{100.53}{ 4\pi})}\sqrt[3]{\dfrac{100.53}{ 4\pi}}=2\sqrt[3]{\dfrac{100.53}{ 4\pi}}$$

$$\approx4.000(m)$$

The height of the cylindrical tank that requires minimum amount of material used is $4$ m.

C.

$$A=2\pi r^2+2\pi rh=2\pi r(r+h)$$

$$=2\pi(2)(2+4)=24\pi(m^2)$$

$$2A=48\pi \ m^2$$

$$\dfrac{48\pi}{28}\approx5.4$$

6 gallons are needed to paint the tank using two coats of paint.