1. ## Optimization Problems

Alright, so i've been doing fairly well in Calculus so far this year, but for some reason optimization problems have been a bit over my head. Here's two that have been trouble for me:

1. Jasper has 100 ft. of fencing to construct a rectangular dog pen and he wants to maximize the area. let L be the length of the pen and W be the width.

a.) Write the formula for area in terms of L.
b.) Determine the dimensions of the pen that will maximize the area.
c.) Suppose there is already a wall in place for one side of the pen. If the same amount of fencing is to be used, what dimensions will maximize the area of the pen?

We're supposed to use the derivative to find the extreme value on both of these problems.

and:

2. Jasper gets two dogs and decides the put a dividing fence in the middle of the rectangle. The fence along the outside of the pen is $4 /ft. but the dividing fence is$16 /ft.

a.) Jasper wants to spend $240 on fencing, so what is the maximum area he can enclose? b.) If Jasper decides to enclose 300 sq. ft., what is the minimum cost? Anyone care to help out a man in need? 2. Originally Posted by Fungus Amongus Alright, so i've been doing fairly well in Calculus so far this year, but for some reason optimization problems have been a bit over my head. Here's two that have been trouble for me: 1. Jasper has 100 ft. of fencing to construct a rectangular dog pen and he wants to maximize the area. let L be the length of the pen and W be the width. a.) Write the formula for area in terms of L. b.) Determine the dimensions of the pen that will maximize the area. c.) Suppose there is already a wall in place for one side of the pen. If the same amount of fencing is to be used, what dimensions will maximize the area of the pen? We're supposed to use the derivative to find the extreme value on both of these problems. and: 2. Jasper gets two dogs and decides the put a dividing fence in the middle of the rectangle. The fence along the outside of the pen is$4 /ft. but the dividing fence is $16 /ft. a.) Jasper wants to spend$240 on fencing, so what is the maximum area he can enclose?
b.) If Jasper decides to enclose 300 sq. ft., what is the minimum cost?

Anyone care to help out a man in need?
For optimization problems, you need two equations to work with. Those equations will depend on what the question is asking for. In this case, you are working with the perimeter and the area.

The equations for perimeter and area for your problem are:

$P = 2W + 2L$

$A = W\cdot L$

a.) Write the formula for area in terms of L.

We know that 100 ft of fencing is available to make the dog pen, so this will be the length of the perimeter:

$100 = 2W + 2L$

$W = 50 - L$

Now plug this into the area equation and a substitution for W, so your area will be in terms of L.

b.) Determine the dimensions of the pen that will maximize the area.

To find the maximum area that can be created take the derivative of your new area equation.

$A = 50L - L^2$

$A' = 50 - 2L$

To find the maximum length, set your equation equal to zero:
50 - 2L = 0

Solve for L. This will be the max length of your dog pen. Then you can plug the value for L into your perimeter equation to find out what W is and then go from there to get your max area.

c.) Suppose there is already a wall in place for one side of the pen. If the same amount of fencing is to be used, what dimensions will maximize the area of the pen?

Look back at our perimeter equation. This part will alter that equation to account for the fact that there are only 3 sides we need to fence in. So, can you think of how you might alter that equation to get the answer for this last one?
Good luck!!