Application of Derivatives-is this correct

Sep 2008
Ch 2.7 #10

The problem is: consider a rectangle in the xy plane, with corners at (0,0), (a,0), (0,b), (a,b). If (a,b) lies on the graph of the equation y=30-x, find a and b such that the area of the rectangle is maximized.

What I got:
A=xy y=30-x so A=x(30-x) = 30x-x^2

A'=30-2x x=15

Is this correct?
Aug 2007
I always wonder about this sort of problem. It does NOT require calculus. It's great to remember your analytic geometry and check it for yourself.

A = x(30-x)

Let's recognize this as a downward-opening parabola.

Zeros: A(0) = 0 and A(30) = 0

What do we know about the relationship of the vertex and the zeros of a downward-opening parabloa.

Vertex: x = (30 + 0) / 2 = 15

A(15) = 225

Is there still a question? You MUST develop some confidence. Go carefully and confidently. Find a way to check your work. Imagine a different method, as shown here, or write so that you can verify your result. Give up guessing - trust me on this.
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MHF Helper
Apr 2005
If you are asking "Is x= 15 the correct answer", no, it is not!

For one thing, the original problem did not ask about "x". It said "find a and b such that the area of the rectangle is maximized." That is, you should give two values as answers.

It would have been better to use "a" or "b" rather than "x" in the formula.

A= ab and b= 30- a so \(\displaystyle A= a(30- a)= 30- a^2\). Now you can either differentiate or, as TKHunny suggests, complete the square to get a= 15. And then b= 30- a= 15 so the correct answer is "the area of the rectangle will be maximized when a and b are both 15."