problem1

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Max-Min Problems:

Problem #1: Find the dimensions of the rectangle of perimeter P that has the largest area.

Instructions:

In this applet, you may enter a value for the perimeter P, vary X by increasing its dimension with the mouse and see the value of the area A changing.

A (X) = X * Y = X* (P/2 - X)

Solution:

A (X) = X * Y = X * (P/2 - X)

A(X) = X*P/2 -X²

A'(x) = P/2 -2*X = 0 --> x = P/4

A"(x) = -2 for any real x --> A(P/4) is a local max

We know that 0.0 <= x <= P/2, and A(0.0) = A (P/2) = 0.0

Therefore X = P/4 and Y = P/2 - X = P/4 are the dimensions of the rectangle sought.

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Write to: Teresinha Kawasaki