Optimization: Am I starting this right?

[sylas7306]

Solved

A farmer wishes to build two pens that share a side. He wishes for the total area of the pen to be 100 square meters. Find the minimum perimeter it will take to construct the pen.

What I have:
Target Equation - P=4x+3y
Constraint - 100=2xy

I take the constraint and solve it for y, which is 50x.
Plug 50x back into the Target Equation: 4x+3(50x)=154x
Take the derivative: 154

I'm not sure what to do next.

 

[stapel]

A farmer wishes to build two pens that share a side. He wishes for the total area of the pen to be 100 square meters. Find the minimum perimeter it will take to construct the pen.

What I have:
Target Equation - P=4x+3y
Constraint - 100=2xy

For what do "x" and "y" stand?

Try using variables that suggest what they stand for. The pens have a length and a width; it appears that we're supposed to assume that the pens are of the same size. So there's a rectangular area with width "w" and length "L", with an extra "L" down the middle to divide the large rectangle into two smaller (but equal-sized) pens. Then each of the pens has width "w/2" and length "L". (Draw the picture, if you're not sure.)

Then the area is the usual A = Lw. But the "perimeter" is actually the length of fencing, which includes that extra length "L" down the middle: P = 2w + 3L. You are given the area: A = 100. So Lw = 100.

Solve the area equation for one of the variables. (It doesn't matter which one.) Plug this into the perimeter equation, and simplify to get the perimeter stated in terms of just one variable. Then differentiate and minimize.

 

[sylas7306]

For what do "x" and "y" stand?

Try using variables that suggest what they stand for. The pens have a length and a width; it appears that we're supposed to assume that the pens are of the same size. So there's a rectangular area with width "w" and length "L", with an extra "L" down the middle to divide the large rectangle into two smaller (but equal-sized) pens. Then each of the pens has width "w/2" and length "L". (Draw the picture, if you're not sure.)

Then the area is the usual A = Lw. But the "perimeter" is actually the length of fencing, which includes that extra length "L" down the middle: P = 2w + 3L. You are given the area: A = 100. So Lw = 100.

Solve the area equation for one of the variables. (It doesn't matter which one.) Plug this into the perimeter equation, and simplify to get the perimeter stated in terms of just one variable. Then differentiate and minimize.

Thank you so much!

 

[HallsofIvy]

You are, of course, welcome to use "x" and "y" as variables but, even if you use "l" and "w", say what they mean!

Here you could say "x is the length of the shared side and y is the length of the other side". Then you have a rectangle with one side of length x and the other of length 2y so area 2xy= 100. Two of the sides have length y and three have length x so the total length of fencing is 2y+ 3x.

Solving the constraint, 2xy= 100, for y, you get y= 100/(2x)= 50/x, NOT 50x! 2y+ 3x= 100/x+ 3x.