A rectangular area is to be enclosed and divided into thirds. The family has $760 to spend for the fencing material. The outside fence costs $10 per running foot installed, and the dividers cost $20 per running foot installed. What are the dimensions that will maximize the area enclosed?

Answer:[tex]x=\frac{19}{3}\ feet\\y=19\ feet[/tex]Step-by-step explanation:Derivatives and Optimization   We can find where a function f has points of maxima/minima by using the first derivative criteria, i.e. f'=0 and evaluate its critical points. Let's say the rectangular area has dimensions x (height) and y (width) and we are going to split the width into thirds. The area of that rectangle is [tex]A=xy[/tex]The perimeter of that rectangle is the length of the outside fence [tex]P_o=2x+2y[/tex]It costs $10 per feet, so the cost of the outside fence is [tex]C_o=10(2x+2y)=20x+20y[/tex]The dividers cost $20 per foot, and we have assumed we split the width, so each divider has a length of x feet. The cost of the internal dividers is [tex]C_i=20(2x)=40x[/tex]The total cost of the fencing is [tex]C=20x+20y+40x=60x+20y=20(3x+y)[/tex]We know there is a limit of $760 to spend for the fencing material, so [tex]20(3x+y)=760[/tex]Reducing [tex]3x+y=38[/tex]Solving for y [tex]y=38-3x[/tex]The area can be now expressed in terms of x alone [tex]A=xy=x(38-3x)=38x-3x^2[/tex]To find the critical point, and possible maxima/minima, we set A'=0 [tex]A'=38-6x=0[/tex]We find: [tex]x=\frac{19}{3}[/tex]Since [tex]y=38-3x=19[/tex]The second derivative is A''=-6 This means the area is maximum at the critical point The dimensions that maximize the area are: [tex]x=\frac{19}{3}\ feet\\y=19\ feet[/tex]