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Given a triangle with vertices at points $(0, -a), (0, a), (b, 0)$, where $a > 0$, find the maximal area and the dimensions (base and height) of a rectangle that can be contained within the triangle.

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I tried to find a function, differentiate it, and find the maximum. I think I kind of did it right, but since I'm not sure I want to ask for some advice.

I set up $x$ as the length of the base and $y$ as the height, but since $y$ is being divided by the $x$ axis in two parts, the function of the area is:

$$\mbox{Area}=x2y$$

and to find y with the triangle, I saw that we can do a little similar triangle to the bigger triangle on the positive side so

$$ \frac ab = y/{b-x}$$

then

$$y=a-{ax}/{b}$$

plugin that into de area equation gives

$$\mbox{Area}=2x(a-{ax}/b)$$

so the derivative is

$$A'=2a-4{ax/b}$$

solving for A'=0 gives $x=b/2$ so finding the dimensions and the area won't be hard once you get the proper x, but I'm kinda doubtful because I didn't set any constraint for the base "x"

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Completing the square:

Maximize $A:= 2x(-(a/b)x+a)$, where $0 \le x \le b$.

$A= 2x((-a/b)x+a)=$

$ -2(a/b)x^2 +2xa= $

$-2(a/b)(x^2-bx)=$

$ -2(a/b)[(x-b/2)^2-(b/2)^2]=$

$-2(a/b)(x-b/2)^2 + 2(a/b)(b/2)^2.$

Maximal area occurs for

$x=b/2$ (why?) : $A_{max}= (ab)/2.$

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Let $A(0,b),$ $B(b,0),$ $C(0,-a)$ and $KLMN$ be our rectangle, $KL=x$, where $K\in BC$ and $N\in AB$.

Thus, since $\Delta ABC\sim\Delta NBK$, we obtain: $$\frac{NK}{2a}=\frac{b-x}{b}$$ or $$NK=\frac{2a(b-x)}{b}.$$ Id est, by AM-GM $$S_{KLMN}=\frac{2a(b-x)x}{b}\leq\frac{2a\left(\frac{b-x+x}{2}\right)^2}{b}=\frac{ab}{2}.$$ The equality occurs for $b-x=x,$ which says that we got a maximal value.

Now, we see that in the optimal case $KL=\frac{b}{2}$ and $NK=a.$

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  • $\begingroup$ hey thanks!!!! I didn's understand the less or equal part, but I see you got the same results, but your method makes more sense. and also I guess that really gets me out of the doubt that I don't need to specifically set a constraint for x (at least for this problem) $\endgroup$ – rorod8 Apr 10 at 3:51
  • $\begingroup$ It's AM-GM: For positives $p$ and $q$ we have $pq\leq\left(\frac{p+q}{2}\right)^2.$ It's just $(p-q)^2\geq0.$ $\endgroup$ – Michael Rozenberg Apr 10 at 3:55

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