Asked by Jane
A rectangle is bounded by the x-axis and the semicircle y = sqrt(36-x^2). What length and width should the rectangle have so that its area is a maximum?
I understand that 2xy = A and that 4x + 2y = P, but I'm not sure how to solve for a variable to plug back into the equation. I keep getting 6, but then the other variable would be 0.
I understand that 2xy = A and that 4x + 2y = P, but I'm not sure how to solve for a variable to plug back into the equation. I keep getting 6, but then the other variable would be 0.
Answers
Answered by
Steve
we don't care about P.
y = √(36-x^2)
a = 2xy = 2x√(36-x^2)
da/dx = 4(18-x^2)/√(36-x^2)
so, da/dx=0 when 18-x^2=0, or x = √18
now you can find y.
y = √(36-x^2)
a = 2xy = 2x√(36-x^2)
da/dx = 4(18-x^2)/√(36-x^2)
so, da/dx=0 when 18-x^2=0, or x = √18
now you can find y.
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