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January 25, 2015

January 25, 2015

Posted by **John** on Sunday, April 6, 2008 at 5:33pm.

The Blue clothing company makes both shirts and pants. Each shirt gives a profit of $25, while a pair of pants yields $55. Both shirts and pants are made using machines and human labor. A single shirt takes an hour of machine labor and an hour of human labor. A pair of pants requires 3 hours of machine labor and 2 hours of human labor. Each day the company can devote a total of 160 hours of human labor and 200 of machine labor. The company must make at least 15 pairs of pants every day, but never has enough material to make more than 70. The number of shirts made in a day can not exceed 100. Lastly, the number of pants made in a day can never be more than the number of shirts. How many pants and shirts should the company make in a day to maximize profits? What is the maximum profit?

So far i set up the profit equation 25x+55y=Profit and now im stuck on the different restrictions for it i think i have y≤70 x≤100 plz help!

- Math Algebra 2 -
**FredR**, Sunday, April 6, 2008 at 6:02pmhuman hrs = 160, or 2x+6y=160

machine hrs = 200, or 2x+9y=200

minimum of 15 pants, or y>=15

must make more shirts than pants, or

x>=y

the maximum profit is where the equation 25x+55y=profit has a local maximum. One way to determine this is using the 1st derivative and set this equal to zero.

- Math Algebra 2 -
**John**, Sunday, April 6, 2008 at 8:39pmi got 18.18 for x and 18.18 for y and the profit would be $1454.54 does that look like its right?

- Math Algebra 2 -
- Math Algebra 2 -
**John**, Sunday, April 6, 2008 at 6:52pmhey thanks alot for all that help i graphed it on some paper and in my calculator and i think for the answer i got 18.18 for x and 18.18 for y and the profit would be $1454.54 can anyone confirm if this is correct? thanks

x=shirt y=pants

- Math Algebra 2 -
**FredR**, Sunday, April 6, 2008 at 11:14pmI got a maximum profit of $4200 with 40 pants and 80 shirts.

- Math Algebra 2 -

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