minimize z = 5x + 15y subject to 4x + 3y <= 72, ....

[Navyguy]

Find the minimum value of z = 5x + 15y, subject to:

. . .4x + 3y <= 72
. . .6x + 10y >= 174
. . .x > 0
. . .y > 0

My answer is (x, y) = (9, 12), but I'm not sure if this is correct. Thank you for checking!

 

[soroban]

Re: minimum value, help please

Hello, Navyguy!

Your answer is correct . . .

Find the minimum value of \(\displaystyle z\:=\:5x\,+\,15y\)

subject to: \(\displaystyle \:\begin{Bmatrix}4x\,+\,3y\:\leq\:72 \\
6x\,+\,10y\:\geq \:174 \\
x\:\geq\:0 \\
y\:\geq\:0 \end{Bmatrix}\)

Same game plan: .(1) Graph the region, (2) Test the vertices.


Since \(\displaystyle x\,\geq\,0,\:y\,\geq\,0\), we are in Quadrant I.


To graph \(\displaystyle 4x\,+\,3y \:\leq\:72\) . . .

Graph the line: \(\displaystyle \,4x\,+\,4y\:=\:72\)
. . Intercepts: \(\displaystyle \,(18,\,0),\0,\,24)\)
. . Draw the line, shade the region below the line.


To graph \(\displaystyle 6x\,+\,10y\:\geq\:174\) . . .

Graph the line: \(\displaystyle \,6x\,+\,10y\:=\:174\)
. . Intercepts: \(\displaystyle \,(29,\,0),\0,\,17.4)\)
. . Draw the line, shade the region above the line.


The region looks like this:

        |
      24o
        |:*
        |:::*
   17.4 o:::::*
        |   *:::*
        |       *:* (9,12)
        |           o
        |             * *
        |               *   *
        |                 *     *
        |                   *       *
        |                     *         *
      --+-----------------------*-----------*----
        |                      18          29

The vertices are: \(\displaystyle \,(0,\,24),\;(0,\,17.4),\;(9,\,12)\)


Test the vertices:

\(\displaystyle \begin{array}{ccc}(0,\,24): & z\:=\:5(0)\,+\,15(24) & = & 360 \\
(0,\,17.4): & \; z\:=\:5(0)\,+\,15(17.4) & \;=\; & 261 \\
(9,\,12): & z\:=\:5(9)\,+\,15(12) & = & \underbrace{225}_{\text{min}} \end{array}\)


Therefore, \(\displaystyle x\,=\,9,\:y\,=\,12\,\) produces the minimum \(\displaystyle z\).

 

[Navyguy]

Thanks again for checking my answer

thanks sooooo much