Question

A pig rancher wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle (see the figure below). He has 620 feet of fencing available to complete the job. What is the largest possible total area of the four pens? A=2x + 5y

Answer 1

radar,
I sense that you already have solved this problem. Ami I wrong?

Answer 2

I think you mean amount of wire used = 2x+5y
then 5y=620-2x
5y=(620-2x)
y=124-(2/5)x
A=x(124-(2/5)x=124-(2/5)x^2
dA/dx=124-(4/5)x set to 0
x=155 ft
y=62 ft
pen will use 2(155) ft + 5(62) ft

Answer 3

the pen has 5 y walls and 2 x walls

Answer 4

yes and as I said, 5 62 ft walls and 2 155' walls

Answer 5

not right you have to use the derivative of A= 2x+5y then solve for whats the greatest area im going to attach a picture.

Answer 6

x = width, which is used 5 times to make 4 pens.

(620 - 5x)/2 = length of pens

y = x(620 - 5x)/2

y = 310x - 5/2x²

The maximum area occurs at the vertex , which is at x = -b/(2a) = -310/-5 = 86 feet width

(620- 5x)/2 - (620 - 5(62))/2 = 310/2 = 155 = combined width of all 5 pens

62 x 155 = 9610 ft² <==ANSWER

Answer 7

i got it thanks

Answer 8

Exactlly, what I had

Answer 9

yes but i need greatest volume

Answer 10

area*

Answer 11

Didn't I give you the greatest area 62 x 155? The area would of been 9610 sq ft.
What was the different asnwer?

Answer 12

You are confusing me, I took the derivative of A for area, set it to 0 and got 62 and 155. Using the restraints on the amount of wire and separate pen requirements. What was done different?

Answer 13

"pen will use 2(155) ft + 5(62) ft" you did + instead of x but its all in all right work,thanks

Answer 14

Glad you now understand how to solve these kind of problems. Good luck with your study efforts.

Answer 15

radar,
It looks like everything worked out fine.
Have a good day.