Asked by Labub
So the q is; You need to enclose a rectangular field with area 9600m^2, then divide the field into 2 sections. Whats the minimal amount of fencing required to accomplish this. So i have 3x+2y, and 9600=xy. i TRIED ISOLATING the second formula and plugging it into the original and finding the derivative. However I keep getting a negative answer. Someone please help
Answers
Answered by
Steve
If f(x,y) is the amount of fence, then
f = 3x+2y = 3x+2(9600/x) = 3x + 19200/x
df/dx = 3 - 19200/x^2
df/dx=0 at x = ±80
f has a max at x = -80, and a min at x=80.
Too bad you didn't show <u>your</u> work ...
f = 3x+2y = 3x+2(9600/x) = 3x + 19200/x
df/dx = 3 - 19200/x^2
df/dx=0 at x = ±80
f has a max at x = -80, and a min at x=80.
Too bad you didn't show <u>your</u> work ...
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