Afeasible region has vertices at (-1,3), (3,5), (4,-1), and (-1,-2) find the maximum and minimum values of each function
f (x, y)=-2x-u
a. max: f (4,-1)=5; min: f (-1,3)=-4
b. max: f (3,5)=19; min: f (-1,-2)=-7
c. max: f (3,5)=13; min: f (4,-1)=-7
d. max: f (-1,-2)=4; min: f(3,5)=-11

Choice D. 
Max: f (-1,-2)=4; min:f(3,5)=-11




Work Shown:

Plug in (x,y) = (-1,3)
f(x,y) = -2x-y
f(-1,3) = -2*(-1)-3
f(-1,3) = 2-3
f(-1,3) = -1

Plug in (x,y) = (3,5)
f(x,y) = -2x-y
f(3,5) = -2*3-5
f(3,5) = -6-5
f(3,5) = -11

Plug in (x,y) = (4,-1)
f(x,y) = -2x-y
f(4,-1) = -2*4-(-1)
f(4,-1) = -8+1
f(4,-1) = -7

Plug in (x,y) = (-1,-2)
f(x,y) = -2x-y
f(-1,-2) = -2*(-1)-(-2)
f(-1,-2) = 2+2
f(-1,-2) = 4

The four outputs are: -1, -11, -7, and 4

The largest output is 4 and that happens when (x,y) = (-1,-2)
So the max is f(x,y) = 4

The smallest output is -11 and that happens when (x,y) = (3,5)
So the min is f(x,y) = -11

This all points to choice D being the answer.

A. p = 2(l + w)p/2 - l = ww = p/2 - l b. while isolating the variable w in the equation, i first divided both sides by 2 and then i subtracted l from both sides. i hope this .