Find the dimensions of a rectangle with perimeter 68 m whose area is as large as possible. (Give your answers in increasing order.)
How do I do this and what are the answers?
Thanks!
2 Answers

Let the sides of the rectangle be length x and y and the area be A
Then 2x+2y=68
So x+y=34
y=34x
A=xy
Substitute the value of y into the equation for A
A=x(34x)
A=34xx^2
dA/dx=342x
Maximum occurs at dA/dx=0
So 34 – 2x = 0
2x=34
x=17
since x+y = 34, y=34
So the largest rectangle with perimeter 68m is the square with sides of length 17m

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