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]
