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mathstresser
10-01-2006, 05:43 PM
Find the absolute maximum and minimum values of of on set D.

f(x, y) = 1 + 4x - 5

D is the closed triangular section with vertices (0, 0), (2, 0), and (0,3 ).

I start by doing the partial derivatives

fx(x, y) = 4
fy(x, y) = -5

So my point is (4, -5), but that is not within the region, so what am I doing wrong?

arthur ohlsten
10-01-2006, 09:04 PM
f[xy]= 4x-4 is a straight line slope 4 y intercept -4

domain D is a right triangle base y=0 0<x<2
"side" x=0 0<y<3
hypoteneuse y=-3/2 x +3 0<x<2 0<y<3

maxis where the line y=4x-4 crosses the hypoteneuse
minimum where the line y=4x-4 crosses the base

I think this is what you need

galactus
10-01-2006, 09:08 PM
I think you have a typo in your function. I am assuming you mean 5y, not just 5. OK?.


Your point lies outside the boundary.

f(x,y)=1+4x-5y....[1]

Let's try some others that lie on the boundary of the triangle.

The line segment between (0,0) and (2,0):

On this line segment we have y=0, so [1] simplifies to a function of a single variable x, f(x,0)=4x+1

This function has no critical points because f'(x)=4 is nonzero for all x.

Thus, the extreme values occur at the endpoints x=0 and x=2, which correspond to (0,0) and (2,0) of D.

The line segment between (0,0) and (0,3):
The same follows with x=0 in our function. The critical values correspond to the endpoints (0,0) and (0,3).

The line segment between (2,0) and (0,3):
The line equation for the line between (2,0) and (0,3) is y=\frac{-3x}{2}+3

Sub this into [1]: 1+4x-5(\frac{-3x}{2}+3)=\frac{23x}{2}+16

This is also nonzero for all values of x, f&#39;(x)=\frac{23}{2}

Therefore, the critical points occur at (2,0) and (3,0).

Put these points in a chart to see which are the min and the max.




(x,y)| (0,0) (2,0) (0,3)
----------------------------------------
f(x,y)| 1 9 -14




The max is at (2,0) and the min is at (0,3).

I hope I didn't overlook something. Check it out.

mathstresser
10-02-2006, 12:33 AM
Thanks!