optimization

md_l_Lopez

New member
Joined
Jan 11, 2022
Messages
2
Barbara (her friends call her Barb because she solves problems about wire) is planning a sustainable fish farm venture for her large property. She has a total of 558 meters of wire for the electrified fencing which will be used to enclose two unconnected pool areas. One of the areas must be square, while the other must be a rectangle that is two times as long as it is wide. The square pond enclosure must contain at least 144 square meters and the other one (the rectangle) at least 1152 square meters. Find the maximum total area of the two enclosures.

I got A=x^2+l*w --> A=x^2+2w^2

Then I use the perimeter equation
558=4x+2w+2l --> 558 = 4x + 6w
w = 93 - 2/3x

Then I plugged it into the area equation
A= x^2+2(2/3x)^2

Then I found the derivative
A(prime) = 34/9x-248

what do I do next?
 
You have a derivative, and are trying to find a maximum. How do those two puzzle pieces fit together?
 
Barbara (her friends call her Barb because she solves problems about wire) is planning a sustainable fish farm venture for her large property. She has a total of 558 meters of wire for the electrified fencing which will be used to enclose two unconnected pool areas. One of the areas must be square, while the other must be a rectangle that is two times as long as it is wide. The square pond enclosure must contain at least 144 square meters and the other one (the rectangle) at least 1152 square meters. Find the maximum total area of the two enclosures.

I got A=x^2+l*w --> A=x^2+2w^2

Then I use the perimeter equation
558=4x+2w+2l --> 558 = 4x + 6w
w = 93 - 2/3x

Then I plugged it into the area equation
A= x^2+2(2/3x)^2

Then I found the derivative
A(prime) = 34/9x-248

what do I do next?
A= x^2+2(2/3x)^2
How did you get that? Please show intermediate steps>
 
What are you studying? Any mention of multivariate calculus or LaGrangian multipliers?

In any case, you have not set things up correctly. What is unknown?

[math]x = \text {side of square;}\\ y = \text {length of rectangle;}\\ w = \text {width of rectangle;}\\ P = \text {combined perimeter;}\\ s = \text {area of square;}\\ r = \text {area of rectangle; and}\\ A = \text {combined area.}[/math]
We have five equations, which can be simplified to four.

[math]A = s + r;\ s = x^2, r = wy, \\ P = 4x + 2y + 2w, \text { and } w = 2y \implies\\ r = 2y^2, \ s = x^2,\ A = x^2 + 2y^2 \text { and } P = 4x + 6y.[/math]
We have three constraints.

[math]r \ge 1152,\ s \ge 144,\ \text { and } P \le 558 \implies \\ 2y^2 \ge 1152,\ x^2 \ge 144, \text { and } 4x + 6y.[/math]
Translating a problem into mathematical notation is something you learned in algebra. You must not forget all of your algebra when you learn calculus. Be careful and orderly in your set up.

OK. What method or methods have you learned for doing optimization with multiple variables subject to constraints?
 
Top