## Calculus Tutorial

#### Intro

This is another hot topic that will absolutely have one or more problems on your calculus exam. For this type of problem first you’ll need an equation of height in terms of radius, so that you can end up with only a variable of r for the equation of minimum cost. Next is to calculate the minimum total cost by adding total cost of bottom, top, and side of the cylindrical container. Finally, differentiate the total cost equation by taking the first derivative of radius r. You will then use the solutions of r to find h and the minimum cost.

#### Sample Problem

A cylindrical can is to hold 20 m^3 The material for the top and bottom costs $10/m^2 and material for the side costs$8/m^2 Find the radius r and height h of the most economical can.

#### Solution

1. Total volume of cylinder: 20 = (area of base)*height = ( r^2)h, we need to solve h in terms of r, so h = 20/r^2

2. List the minimum equation for total cost of construction of cylinder:
C = (total cost of bottom) + (total cost of top) + (total cost of side)
= (unit cost of bottom)(area of bottom) + (unit cost of top)(area of top) + (unit cost of side)(area of side)
= $10( r^2) +$10( r^2) + $8(2 rh) = 20 r^2 + 16 rh 3. Substitute h with 20/r^2 in step 1 to cost equ in step 2: C = 20 r^2 + 320 /r 4. Now differentiate the C equ, we get C’ = 40 r +320(-1/r^2) = 40 r[r^2/r^2] – 320 /r^2 = 40 (r^3-8)/r^2 = 0, so solve for r and we get r = 2. Then h = 20/2^2 = 20/4 = 5 5. Calculate C by plugging r and h values from #4: C =$240 = \$754(ans) 