Monday

March 30, 2015

March 30, 2015

Posted by **Anonymous PLEASE HELP!** on Monday, March 4, 2013 at 11:12am.

If each scientific calculator sold results in a $2 loss,but each graphing calculator produces a $5 profit, how many if each type should be made daily to maximize net profits?

A) Define the objective equation.

B)Define the constraint equations.

C)Graph the associated constraint equations.

D) Determine the vertices of each of the graphed equations.

E) Determine how many of each type of calculators should be made daily to maximize the net profits and the amount of net profit.

- College Algebra -
**Steve**, Monday, March 4, 2013 at 11:59ammaximize p = -2x+5y subject to

x <= 200

y <= 170

x+y >= 200

looks like 30 scientific, 170 graphing

- College Algebra -
**Anonymous PLEASE HELP!**, Monday, March 4, 2013 at 12:18pmHow did you get 30 and 170? If the equation is -2x+5y?

- College Algebra -
**Steve**, Monday, March 4, 2013 at 2:20pmWell, just intuitively, if you lose money on each scientific calculator, you want to produce as few as possible. Ideally, zero.

Unfortunately, you have to produce at least 200 boxes a day, and can only produce 170 graphing calculators. So, you gotta do at least 30 of the losers.

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