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I am working on finding the largest rectangle inscribed in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Then its given (this is an example, so answer given)

Maximize $A(x,y) = 4xy$.

$$f_x=\lambda g_x \rightarrow 4y = \frac{2\lambda x}{a^2}$$

$$f_y = \lambda g_y \rightarrow 4x = \frac{2 \lambda y}{b^2}$$

$$\frac{y}{x} = \frac{xb^2}{ya^2} \rightarrow \frac{x}{y} = \frac{a}{b}$$

Then question is why in the next step:

$$x=ak,\qquad y=bk?$$

Where did $a, b$ come from? Why isit not $x=a, y=b$?

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x/y=a/b is not the same as x=a and y=b. Consider 1/1=2/2, yet 1$\ne$2. –  anon Apr 8 '12 at 3:16

1 Answer 1

up vote 4 down vote accepted

You cannot conclude $x=a$ and $y=b$ because two fractions can be equal without being identical. For instance, $$\frac{3}{6}\text{ is equal to }\frac{1}{2},\text{ but we do not have }3=1\text{ and }6=2.$$

However, you can conclude from $\frac{x}{y} = \frac{a}{b}$ that $x=\frac{y}{b}a$; and $y =\frac{bx}{a} = \frac{x}{a}b$. However, $\frac{y}{b} = \frac{x}{a}$ (since $\frac{x}{y}=\frac{a}{b}$), so we can set $k=\frac{y}{b}=\frac{x}{a}$ to conclude that $x=ka$ and $y=kb$.

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