The minimum value of z=4x+10y subject to

3x+y<=to24

6x+4y<=to66

x>=to0,y>=to 0. is.

My answer after looking at a list of possible answers is 32, is this right.

a.165

b.110

c.44

d.32

Thanks for any help given.

Results 1 to 11 of 11

- Jul 28th 2006, 01:01 AM #1

- Joined
- Jun 2006
- Posts
- 165

- Jul 28th 2006, 04:13 AM #2

- Jul 28th 2006, 02:07 PM #3

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 12,028
- Thanks
- 848

Hello, kwtolley!

Am I reading it wrong?

Are there typos in the problem?

The minimum value of $\displaystyle z \:=\:4x+10y$ subject to: $\displaystyle \begin{Bmatrix}3x+y\,\leq\,24 \\ 6x+4y\,\leq\,66 \\ x \geq 0\\ y \geq 0\end{Bmatrix}$ is:

$\displaystyle a)\;165\qquad b)\;110\qquad c)\;44\qquad d)\;32$

Doesn't $\displaystyle x = 0,\;y = 0$ satisfy the inequalties?

- Jul 28th 2006, 03:01 PM #4

- Jul 28th 2006, 03:13 PM #5
Soroban, I just ran this through the Linear Programming solver in Maple and it gave me (0,0) as the minimum. I looked right past the origin.

Looks like the list of answers is omitting the correct minimum. That's why you're the man and I'm just a dogface .

Good 'ketch' .

- Jul 29th 2006, 01:20 AM #6

- Joined
- Jun 2006
- Posts
- 165

- Jul 29th 2006, 05:18 PM #7

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- Jul 29th 2006, 05:36 PM #8

- Jul 29th 2006, 06:04 PM #9

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- Jul 29th 2006, 07:19 PM #10Originally Posted by
**kwtolley***minimum*value is 0, not on the list. So there is a typo. From this linear programming solver, which you can use to both solve the problem or check your work, the*maximum*value is 165, answer a.

If you have linear programming problems to solve in the future, you may want to bookmark that solver.

- Jul 30th 2006, 01:50 AM #11