Post #37

AP this week =0.

I'll explain optimization because I am presenting it tomorrow.
Steps:
1. Determine everything you are given. Determine your primary and secondary equation. Your primary equation is the one you are maximizing or minimizing. Your secondary equation is usually using the other information given to you in the problem.

2. Solve the secondary equation for one variable

3. Plug the solved secondary equation into the primary equation and simplify.

4. Take the derivative of that equation.

5. Solve that equation for the remaining variable.

6. Plug that variable back into the secondary equation to find the other variable(s).

Example (some of you will see this problem on the white board tomorrow):
We want to construct a box whose base length is 3 times the base width. The material used to build the top and bottom costs $10/ft^2 and the material used to build the sides cost $6/ft^2. If the box must have a volume of 50ft^3 determine the dimensions that will minimize the cost to build the box.

1. You are looking for the dimensions to minimize the box so your primary equation will be
c= 10(2lw) + 6(2wh + 2lh)
Note: you multiply by 10 because of the $10/ft^s for the top and bottom, and 6 for the $6/ft^2 for the sides.
Once distributed in, that equation simplifies to 60w^2+48wh
Your secondary equation will deal with volume since you are given the restraint of v=50ft^3
The volume of a box is lwh so 50 = lwh
Since l is 3 times the width, l=3w
The secondary equation is 50 = 3w^2h

2. Solve for h: h=50/3w^2

3. Plug into primary: 60w^2 + 48w (50/3w^2)
Simplified is 60w^2 + 800/w

4. Derivative: 120w - 800w^-2
which equals: 120w^3 - 800/ w^2

5. Solve for w by setting the top of the equation equal to 0.
120w^3 -800 = 0
w= 1.8821

6. Plug into equations to find other 2 variables:
50 = 3w^2 h
50= 3(1.8821)^2 h
h=4.7050

l=3w
l=3(1.8821)
l= 5.6463

Hope this helps.

I have questions on the invertible problems and the related rate problems with surface area such as number 13 on one of the last aps we took.