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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 $0$.

I don't know where I am making a mistake, so please help me. Thank you in advance.

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    $\begingroup$ 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. $\endgroup$ – Théophile Jul 3 at 15:38
  • $\begingroup$ 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. $\endgroup$ – farruhota Jul 3 at 16:06
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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 $|x-1|$.

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    $\begingroup$ Thank you so much. After I saw your answer I used a graph calculator and x-1 makes a lot of sense :D $\endgroup$ – curiouseng Jul 3 at 15:40
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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 $c(1)^{3} = c$.

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