Okay, so I understand that the equation is a downwards facing parabola with a y-intercept at 12. I don't understand what it means by upper vertices? I know that the answer is 32 but I don't understand how to get there. Can someone please guide me and explain to me the process of solving this problem? Thank you!
[Math] A rectangle with base on the x-axis has its upper vertices on the curve y=12-x^2 . Find the maximum area of such a rectangle.
calculus
Best Answer
Hint: Draw a picture of the downward-facing parabola, and of a rectangle of the type described.
Let $(x,y)$ be the upper right-hand corner of the rectangle. Then by symmetry the base of the rectangle has length $2x$, and the height is $y$, that is, $12-x^2$.
So the area $A(x)$ of the rectangle is given by $$A(x)=2x(12-x^2).$$ Maximize, using the usual tools. Note that we must have $0\le x\le \sqrt{12}$.