1. ## Variation Worded Problem

The price of painting the outside of a cylindrical tank (the bottom and top are not painted) of radius r and height h varies directly as the total surface area. If r=5 and h=4, the price is $60. What is the price when r=4 and h=6 2. Originally Posted by Meg.Fondale The price of painting the outside of a cylindrical tank (the bottom and top are not painted) of radius r and height h varies directly as the total surface area. If r=5 and h=4, the price is$60.
What is the price when r=4 and h=6
If you open up a cylinder into a net then you will observe that it is a rectangle with the top and the buttom being a circle. As the question says the top and the buttom are not painted then this implies that we only consider the side, which is the rectangle.

The area of a rectangle is width multiplied by length. For the cylinder, the length is the height ( $h$) and the width is the circumference (because this is the length of the circle) this it is $2\pi r$.

Therefore, the surface area considered is $S = (2\pi r)(h) = 2 \pi h r$.

When $r=4, h=5$, the surface area is $S = 2 \times \pi \times 5 \times 4 = 40\pi$ at which the price is $\ 60$.

Now consider the surface area when $r=4, h=6$, this would equal $S = 2 \times \pi \times 4 \times 6 = 48\pi$.

Now we are able to use general arithmetic (Ratio consideration) to solve for the price. When $r=4, h=6$, therefore the price ( $P$) is $P= \frac{48 \pi \times 60}{40\pi} = ...$

EDIT: I have attached the cylinder diagram and the net considered.

3. Originally Posted by Meg.Fondale
The price of painting the outside of a cylindrical tank (the bottom and top are not painted) of radius r and height h varies directly as the total surface area. If r=5 and h=4, the price is \$60.
What is the price when r=4 and h=6
To learn how to set this up as a variation equation, try here.

Once you have learned the basic terms and techniques, the following should make sense:

i) The variation equation must be of the form P = k(SA), where P is the price, k is the constant of variation, and SA is the surface area.

ii) You know that SA = 2(pi)rh, so P = 2k(pi)rh.

iii) You are given that, for r = 5 and h = 4, P = 60. Use this to solve for the constant of variation k.

iv) Now that you have the value of k, plug this into P = 2k(pi)rh. This is your variation equation.

v) Plug in 4 for r and 6 for h. Simplify to find the required value for P.