SOLUTION: A piece of wire 64 cm long is cut into two pieces with different lengths. The two pieces are formed to TWO squares. (a) Show that the combined area of the two squares A is given

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Question 1103715: A piece of wire 64 cm long is cut into two pieces with different lengths. The two pieces are formed to TWO squares.
(a) Show that the combined area of the two squares A is given by A=2x^2-30x+256
(b) What is the minimum combined area of the two squares?
Answer by math_helper(2461) About Me  (Show Source):
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A piece of wire 64 cm long is cut into two pieces with different lengths. The two pieces are formed to TWO squares.
(a) Show that the combined area of the two squares A is given by A=2x^2-30x+256
(b) What is the minimum combined area of the two squares?
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Let x = the length of the first piece
the 64-x = the length of the 2nd piece
Area of the first square is +%28x%2F4%29%5E2+
Aree of the 2nd square is +%28%2864-x%29%2F4%29%5E2+
Combined area = +%28x%5E2%2F16%29+%2B+%28%284096-128x%2Bx%5E2%29%2F16%29+
This simplifies to +highlight%28matrix%281%2C3%2C+%22+%22%2C+%281%2F8%29%28x%5E2-64x%2B2048%29%2C%22+%22%29%29+
Notice that this disagrees with the value of A in the problem statement. I checked my work but I've been known to make errors.
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To find the minimum, take the derivative. I will use my equation (the process is the same even if the correct equation is different):

+A+=+%281%2F8%29%28x%5E2-64x%2B2048%29+
Take the derivative of A with respect to x:
+dA%2Fdx+=+%281%2F8%29%282x+-+64%29+
Set it to zero:
+++%281%2F8%29%282x-64%29+=+0++ —>
x+=+32cm
Min Area = +%2832%2F4%29%5E2+%2B+%2832%2F4%29%5E2+=+8%5E2+%2B+8%5E2+=+highlight%28128cm%5E2%29+
This is a local minimum of A because +d%5E2A%2Fdx%5E2+=+%281%2F8%29%282%29+ —> concave up. Normally, you should also check the endpoints (here x=0, x=64) but due to the quadratic form, we know the local minimum is a global minimum on the interval.