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I've been working on a problem set of math problems lately, but I've come across some difficulty with the problem below.

The dimensions of a rectangular piece of paper ABCD are AB=10 and BC=9. It is folded so that corner D is matched with a point F on edge BC. The lengths of EF, EC, and FC are all determined by the length of DE. Let DE=x.
a. Write and equation for the area of EFC in terms of x.
b. Find the value for x that maximizes the area of EFC.

The given illustration.

I understand that we could use the Pythagorean theorem, but I'm not quite sure how to make it work. Additionally, if the answer involves calculus, inform me, as I will be unable to work with that due to my lack of understanding of that subject matter. Also, providing the equations for it would be especially helpful.

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From your diagram, $EC = DC - DE$

Since it's a rectangle, $DC = AB = 10$. $DE = x$ is given.

$EF = DE$ since $EF$ is just the folded version of $DE$

You have two sides of the triangle $EFC$, so you can calculate the third side using Pythagorean's theorem.

The area of a triangle is $\frac{1}{2} base * height$. Since you know the base and the height, you can calculate this.

To maximize the area, take the derivative of your formula for the area with respect to x, set to 0 and solve for x. This part is calculus.

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  • $\begingroup$ Is there any way to determine the later without using a derivative? $\endgroup$ – Aidan Boyle Sep 8 '17 at 3:09
  • $\begingroup$ Yeah there are a few ways but they would be specific to the situation. It should be clearer once you have the equation of the area in terms of x. For instance, if the area is an upside down parabola, like $-x^2 + 4$ then the maximum is at 4.This is because -x^2 is always 0 or a negative number. $\endgroup$ – FullofDill Sep 8 '17 at 3:16
  • $\begingroup$ Also, could you try to provide an equation for this problem? My attempts usually result in some form of imaginary number, and I wanted to verify its correctness. $\endgroup$ – Aidan Boyle Sep 8 '17 at 4:42
  • $\begingroup$ It will be slightly easier if we start from letting EC = y instead and figure out the x later. $\endgroup$ – Mick Sep 8 '17 at 8:41
  • $\begingroup$ @aiden Boyle if you edit your question to show your attempt, I can explain where your mistake is. There won't be any imaginary numbers if you follow the steps provided in my answer $\endgroup$ – FullofDill Sep 8 '17 at 11:14

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