Consider the region R in the 1st quadrant bounded on the left by $y=x^2$, on the right by $y=(x-5)^2$ and below by the x-axis. Find the area of the largest rectangle inscribed inside region $R$.
I have attempted this question, though I don't know the correct solution. I subtracted both equations for the length, but was having trouble deciding on the expression for the height. I settled on $x^2$ as the expression for height, my answer is $A=23.14$
Best Answer
the region bounded on the left by $y = x^2$ on the right by $y = (5-x)^2$ and below by $y = 0$ is symmetric about $x = 5/2.$ the width of the rectangle is $2(5/2 - x) = 5 - 2x$ and the height is $x^2$ so the area to maximize is $$A= (5-2x)x^2, 0 \le x \le 5/2$$ the critical numbers of $A$ are $0, 5/3$ evaluating $A$ at the boundaries $x = 0, x = 5/2$ and at the critical number the maximum area is $A(5/3)=125/27.$ i hope i did not make any arithmetical errors.