David wants to build a rectangular fencing with the 5 identical parts for his animals. He has 780 feet of fencing to make it. What dimensions of each part will maximize the total enclosed area?

David wants to build a rectangular fencing with the 5 identical parts for his animals He has 780 feet of fencing to make it What dimensions of each part will ma class=

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frika

Answer:

65 ft by 39 ft

Step-by-step explanation:

Let x ft be the length of each part and y ft be the width of each part. The total perimeter of all 5 parts is

[tex]6x+10y=780[/tex]

The length of the large rectangle is 5y ft and the width of the large rectangle is x, so the total area is

[tex]A=x\cdot 5y[/tex]

From the first equation,

[tex]y=78-0.6x[/tex]

Substitute it into the area expression:

[tex]A(x)=x\cdot 5(78-0.6x)=390x-3x^2[/tex]

Find the derivative:

[tex]A'(x)=390-3\cdot 2x=390-6x[/tex]

Equate it to 0:

[tex]390-6x=0\\ \\6x=390\\ \\x=65[/tex]

Then

[tex]y=78-0.6\cdot 65=78-39=39[/tex]