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Topic: Trigonometric area optimization
Replies: 13   Last Post: Dec 20, 2012 3:29 PM

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Peter Scales

Posts: 122
From: Australia
Registered: 4/3/05
Re: Trigonometric area optimization
Posted: Dec 11, 2012 12:25 PM
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Hi Adam,
>
> The lines y=10-2x, y=mx and y=(-1/m)x, where m > 1/2,
> form a right triangle. Find an m, so that the area
> measurement of the triangle is as small as possible.
>
>

Let line y=m.x be OA, line y=(-1/m).x be OB
and line y=10-2.x be AB

Then Area =1/2 .OA .OB

A is 10/(m+2),m.10/(m+2)
B is 10/(2-1/m),10/(1-2.m)

OA=sqrt(1+m^2).10/(m+2)
OB=sqrt(1+m^2).10/(2.m-1)

Area=1/2 .OA .OB =50.(1+m^2)/(2.m^2+3.m-2)
Find dArea/dm and set it to = 0
and select the option with m>1/2

Sketch the result and check Area for m=0 and
a few other values of m, including both solutions
of the quadratic dArea/dm=0.

Regards, Peter Scales.



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