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radar, I sense that you already have solved this problem. Ami I wrong?
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
the pen has 5 y walls and 2 x walls
yes and as I said, 5 62 ft walls and 2 155' walls
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.
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
i got it thanks
Exactlly, what I had
yes but i need greatest volume
Didn't I give you the greatest area 62 x 155? The area would of been 9610 sq ft. What was the different asnwer?
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?
"pen will use 2(155) ft + 5(62) ft" you did + instead of x but its all in all right work,thanks
Glad you now understand how to solve these kind of problems. Good luck with your study efforts.
radar, It looks like everything worked out fine. Have a good day.