# If $y = cx^2$ for $c>0$ and $0 \leq x \leq 1$, what is the largest possible area of the rectangle?

Consider the collection of rectangles with a vertex at $$(1,0)$$ and the other lying on $$y = cx^2$$ for some $$c > 0$$ and $$0 \leq x \leq 1$$. Find the rectangle with maximal area.

So I named the height as $$y$$ and width as $$x$$ and the area formula became $$A = xy$$. Then I use the given equation $$y=cx^2$$ to sub in y in the area formula and I got $$A = x\cdot cx^2$$ which became $$A = cx^3$$. Then I took the derivative to find the max area and I got $$A' = 3cx^2$$. When I tried to make it equal to $$0$$, only critical point is $$x = 0$$ and it doesn't make sense since a rectangle's width cannot be $$0$$.

• A comment on your reasoning: overlooking the error with the width (as covered in ryszard eggink's answer), note that your conclusion is wrong. Since $x$ is bounded, you should check the boundaries too, i.e., $x=0$ (which is already a critical point) and $x=1$. At $x=1$, we would in principle have a maximum area with $A=c1^3=c$, and this would be the solution. At $x=0$, we would have a minimum area. Furthermore, instead of saying that a rectangle can't have width $0$, it's more informative to see this as a degenerate case: the rectangle is so thin that it disappears entirely. – Théophile Jul 3 at 15:38
• has it been taken from a textbook? some more restrictions are needed: 1) the other vertex can be the adjacent vertex or opposite vertex; 2) the sides of rectangle can be parallel to axes or not. – farruhota Jul 3 at 16:06

I assume that sides of the rectangle are parallel to axis. Then the desired area is:

$$A(x)=cx^2|x-1|=cx^2(1-x)$$ The equality goes from the fact that $$x\in(0,1)$$.

Then taking the derivative $$A'(x)=2cx(1-x)-cx^2$$

The derivative is zero iff $$2cx(1-x)-cx^2=cx(2-3x)=0$$ Then for $$x=\frac{2}{3}$$ the area is maximal (you can take the second derivative to see that it is indeed a maximum).

EDIT: To your solution, you mistakenly took the length of the $$x$$-axis side to be $$x$$, but it is $$|x-1|$$.

• Thank you so much. After I saw your answer I used a graph calculator and x-1 makes a lot of sense :D – curiouseng Jul 3 at 15:40

The absolute maximum of a continuous function over a closed and bounded interval will occur either at a critical value in that interval or at an endpoint. In this case, since the critical point $$x = 0$$ (which is also an endpoint) is clearly not the answer, check the other endpoint $$x = 1$$, which in this case gives the maximum area $$c(1)^{3} = c$$.