View Full Version : absolute max/min of f(x, y) = 1 + 4x - 5 on set D

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

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.


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.

10-02-2006, 12:33 AM