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November 22, 2014

November 22, 2014

Posted by **Sribava** on Sunday, November 6, 2011 at 3:21pm.

- calculus -
**Steve**, Monday, November 7, 2011 at 5:29amInteresting problem. Check to make sure I get this right.

So, we have a triangle with base k, and vertex where the line through (1,1) and (0,k) intersects y=3x.

The line through (1,1) and (0,k) is

(y-1)/(x-1) = -1/(k-1)

y = (x-1)/(1-k) + 1

So, that line intersects y=3x when

3x = (x-1)/(1-k) + 1

x = k/(3k-2)

y = 3k/(3k-2)

The area a of the triangle is thus

a = 1/2 k * 3k/(3k-2) = 3k^2/(6k-4)

da/dk = (6k(6k-4) - 3k^2 * 6)/(6k-4)^2

= 6k(3k-4)/(6k-4)^2

da/dk = 0 when k=0 or k = 4/3

Thus, the vertex of the triangle is where the line through (1,1) and (4/3,0) intersects y=3x

(y-1)/(x-1) = 1/(-1/3)

y = -3x + 4

That line has slope -3

. . .

-3x + 4 = 3x

6x = 4

x = 2/3

y=2

So, the triangle has base k=4/3, height h=2, area a=4/3

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