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Find equation of line passing through $(20,12)$ such that the area of the triangle formed by the line and the positive axis is smallest possible.

Also: $\frac{x}{a}+\frac{x}{b}=1$
where $a, b$ are x, y intercepts

So far, I have

$A=\frac{1}{2}ab, \qquad line = (20,12)+\lambda(a,-b)$

I need to find min area, so I need to differentiate $A$ but with respect to what? From what I have, I can't really find a way to express 1 term in terms of another. If I try to express 1 term in terms of 3 others from $\frac{x}{a}+\frac{x}{b}=1$, it doesn't appear useful.

UPDATE

I think the picture will look like:

enter image description here

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    Yes thats what I mean, see update for a picture2012-03-18

1 Answers 1

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(This is off the top of my head, so MMMV.)

Since the line passes through (20, 12), $20/a + 12/b = 1$, or $20/a = 1-12/b = (b-12)/b$, or $a = 20b/(b-12)$, so $ab/2 = 20b^2/(b-12)$.

The derivative, ignoring the 20, is $\frac{b^2 - 2b(b-12)}{(b-12)^2}$ which is zero when $b^2 = 2b(b-12)$ or $b = 2(b-12)$ or $b=24$.

$a$ then is $20\ 24/12$ or $40$, and the area is 480.

As to why this lists differential equations and vector spaces, beats me.

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    MMMV = My Mileage May Vary (my punning takeo$f$f on YMMV).2012-03-18