Optimization Problem - # 4

[Jason76]

Mr. Jones wants to separate Jasper and the raccoon. He has 160 feet if fencing to enclose them in 2 adjacent pens. What dimensions should be used so that the enclosed area will be a maximum?

I'm thinking 160 is the parameter. How would 2 adjacent pens fall into this?

If this optimization problem starts with parameter, then you would solve the parameter equation for y. Then plug the whole equation into the area equation, find the derivative of the area equation, solve for a critical number. Plug that critical number back into the parameter equation.

 

[stapel]

Mr. Jones wants to separate Jasper and the raccoon. He has 160 feet if fencing to enclose them in 2 adjacent pens. What dimensions should be used so that the enclosed area will be a maximum?

I'm thinking 160 is the parameter. How would 2 adjacent pens fall into this?

I don't know what you mean by "falling into this" (nor to what "this" you're referring), but this is the same exercise you encountered back in algebra when you did quadratic max/min word problems. You can use the same set-up here.

 

[Jason76]

Goal: Find Maximum Area

I'm assuming were given the perimeter info, so then we solve the perimeter equation for y. We then plug that y value into the area equation. Take the derivative of the area equation, find the critical number. Finally plug that critical number into the area equation. In addition, we use the 2nd derivative test to confirm we are getting what we want (a maximum).

 

[Jason76]

Perimeter -= P

Area = A

\(\displaystyle P = 3x + 2y = 160\)

\(\displaystyle y = 80 - \dfrac{3}{2}x\)

\(\displaystyle A = x(80 - \dfrac{3}{2}x)\)

\(\displaystyle A = 80x - \dfrac{3}{2} x^{2}\)

\(\displaystyle A' = 80 - 3x\)

\(\displaystyle 80 - 3x = 0\)

\(\displaystyle x = \dfrac{80}{3}\)

\(\displaystyle A = (\dfrac{80}{3})(80 - \dfrac{3}{2} x)\) Next move?

 

[Jason76]

Perimeter -= P

Area = A

\(\displaystyle P = 3x + 2y = 160\)

\(\displaystyle 3x + 2y = 160\)

\(\displaystyle 2y = 160 - 3x\)

\(\displaystyle y = 80 - \dfrac{3}{2}x\)

\(\displaystyle A = x(80 - \dfrac{3}{2}x)\)

\(\displaystyle A = 80x - \dfrac{3}{2} x^{2}\)

\(\displaystyle A' = 80 - 3x\)

\(\displaystyle 80 - 3x = 0\)

\(\displaystyle x = \dfrac{80}{3}\)

\(\displaystyle A = (\dfrac{80}{3})(80 - \dfrac{3}{2} x)\) - Looks right to me. See any error?