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Find the maximum area of a rectangle in the xy-plane with its sides parallel to the axes, one vertex at the origin, and the diagonally opposite vertex on the curve $$ x^2 + y = 1 $$

I am supposed to first incorporate the constraint into the formula for area (length times width) but unsure how to find the length and width of this rectangle. Any tips?

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  • $\begingroup$ $x$ and $y$ are length and width. $\endgroup$ – Emanuele Paolini Oct 27 '14 at 22:58
  • $\begingroup$ How is that incorporating the constraint of the curve? $\endgroup$ – vinamrata Oct 27 '14 at 22:58
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Hint

If you assume that the length of basis is $x$ ($0<x<1$) then its weight is $y=1-x^2.$ So, its area is $x\cdot (1-x^2).$ You need to find the maximum of this function.

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  • $\begingroup$ Ah I'm such a doof. I couldn't get the x part. Thank you so much! $\endgroup$ – vinamrata Oct 27 '14 at 23:01
  • $\begingroup$ You're welcome. $\endgroup$ – mfl Oct 27 '14 at 23:44
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You want to maximize $xy$ with the costraints $x\ge 0$, $y\ge 0$, $x^2+y=1$. Solve $x^2+y=1$ for $y$ and replace in the function to be maximized.

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One possible way is using Lagrange Multiplier.

We want to maximize $f(x,y)$ Subject to $g(x,y)=c$

In this case, $f(x,y)=xy$ , $g(x,y)=x^2+y$ and $c=1$

We'll define $L(x,y,\lambda)=f(x,y)+\lambda(g(x,y)-c)=xy+\lambda(x^2+y-1)$
We want to find $x$ and $y$ such that:
$\partial L / \partial x = 0$
$\partial L / \partial y = 0$
$\partial L / \partial \lambda = 0$

From that we get:
$y+2\lambda x=0 \Rightarrow y=-2\lambda x$
$x+\lambda=0 \Rightarrow x=-\lambda$
$x^2+y-1=0 \Rightarrow y=1-x^2$

Solving for $x$ and $y$ (which should be trivial now) yields: $x=1/\sqrt3$ and $y=2/3$ In other words, $f(x,y)=\dfrac{2}{3\sqrt3}=\dfrac{2\sqrt3}{9}$

Just like some of the other answers posted here.
(Solution might seem too complex compared to other solutions, but it is worth learning about Lagrange Multiplier since it comes up often in more advanced topics).

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Let the opposite vertex have coordinates $(x,y)$. Because this point is on the given curve than it must satisfy $$x^2+y=1$$ On the other hand you are interested in the area of the rectangle which happens to be $$A=|xy|$$ So your problem is $$\max_{x,y}\{A=|xy|\}$$ subject to $$x^2+y=1$$ Now there are several ways to approach this problem. One is by using Lagragean method, another method is by substituting the constraint in the objective function. We will follow the second as it is more appropriate in this case i.e. $$\max_{x}\{A=|x(1-x^2)|\}$$ Notice that if $(x^*,y^*)$ is a solution so is $(-x^*,y^*)$ if $x\geq0$ then the first order condition $$\frac{dA}{dx}=1-3x^2=0\Rightarrow x^*=\sqrt{3}/{3}\Rightarrow y^*=2/3$$ if $x<0$ then the first order condition reads $$\frac{dA}{dx}=-1+3x^2=0\Rightarrow x^*=-\frac{\sqrt{3}}{3}\Rightarrow y^*=2/3$$

In both cases the maximum area is $A=\frac{2\sqrt{3}}{9}$.

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