Pre Calc homework need some help

[nate2587 10+ year member]

The future homemakers Club is making canvas tote bags and leather tote bags for a money making project. Both types of tote bags will be lined with canvas and have leather handles. For the canvas tote bag, they need 4 yards of canvas and 1 yard of leather. For the leather tote bags, they need 3 yards of leather and 2yards of canvas. The adviser had purchased 56 yards of leather and 104 yards of canvas. If the clube decides to sell the canvas bags for $20 and the leather bags for $35each, what is the maximum number of bags they need to make for a maximum profit? What is the maximum profit?

^^ i dont understand the question at all... i think its a matrix problem or something

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[Trey803 5,000+ posts]

You gunna need to graph a function on a calculator then find the max.(if i remeber correctly)

 

[nate2587 10+ year member]

we arnt allowed to use a calc and i guess she has ways to check to see if we did it with a calc

even like an idea of where to get started will help

 

[Tiger Bass 5,000+ posts]

Here's the long way because I dont feel like explaining the short way:

First, figure out how many canvas bags you can make when you don't make any leather bags. Then find out how many canvas bags you can make when you make 1 leather bag. Continue that process until you are making no canvas bags and only leather. Then add up the profit for each combination of bags. Whichever one makes you the most money is the one you should pick.

It will be easier if you set up a little chart, starting with the maximum amount of canvas bags, working your way down to the maximum amount of leather bags.

 

[Twistid 5,000+ posts]

hire asians to make them for you

 

[PatFitz9 10+ year member]

there are two options (canvas bags vs. leather bags), and you have the material needed for both. figure out what you need and how much $$$ you make, problem solved.

 

[nate2587 10+ year member]

Here's the long way because I dont feel like explaining the short way:
First, figure out how many canvas bags you can make when you don't make any leather bags. Then find out how many canvas bags you can make when you make 1 leather bag. Continue that process until you are making no canvas bags and only leather. Then add up the profit for each combination of bags. Whichever one makes you the most money is the one you should pick.

It will be easier if you set up a little chart, starting with the maximum amount of canvas bags, working your way down to the maximum amount of leather bags.

alright so i started to work out the graph on paper and how would i setup the equations to figure it out short hand

 

[nate2587 10+ year member]

bump

 

[squeak12 10+ year member]

Set up an equation so you can graph it. Then take the derivative and find the zeros of that equation. Then test to see at which point the values go from positive to negative and that the MAX.

 

[nate2587 10+ year member]

Set up an equation so you can graph it. Then take the derivative and find the zeros of that equation. Then test to see at which point the values go from positive to negative and that the MAX.

i need the equation i dunno what it is

 

[nate2587 10+ year member]

bump

 

[wu36ca 10+ year member]

These are original equations

M = 20X(I) + 35X(II)

104 = 4X(I) + 2X(II)

56 = X(I) + 3X(II)

________________________________

104 = 4X(I) + 2X(II)

-244 = -4X(I) - 12X(II) (this is the second equation multiplied by -4, so we can add the two equations together)

-140 = -10X(II)

14 = X(II)

Then plug X(II) into any of the above equations:

X(I) + 3(14) = 56

X(I) + 42 = 56

X(I) = 14

so M = 20(14) + 35(14) for a maximum profit of $770.

I am such a nerd for even attempting this, but my GF is an even bigger nerd for remembering how to do it. The sad part is I tutored linear programming in college. BUt that was 2 years ago and I do not remember shit.

 

[surftb15 10+ year member]

yea you have to set up a system of equations first, this isnt pre calc, this is more like algebra

 

[nate2587 10+ year member]

These are original equationsM = 20X(I) + 35X(II)

104 = 4X(I) + 2X(II)

56 = X(I) + 3X(II)

________________________________

104 = 4X(I) + 2X(II)

-244 = -4X(I) - 12X(II) (this is the second equation multiplied by -4, so we can add the two equations together)

-140 = -10X(II)

14 = X(II)

Then plug X(II) into any of the above equations:

X(I) + 3(14) = 56

X(I) + 42 = 56

X(I) = 14

so M = 20(14) + 35(14) for a maximum profit of $770.

I am such a nerd for even attempting this, but my GF is an even bigger nerd for remembering how to do it. The sad part is I tutored linear programming in college. BUt that was 2 years ago and I do not remember shit.

well thanks for the info appreciate it. i worked it out the long way and wouldn't the final answer be maximum profit of $820 with 20 canvus bags at 400 dollars profit and 12 leather bags at $420 profit?

 

[nate2587 10+ year member]

The ratio for 1 leather bag is 3 yard leather 2yard canvas

The ration for 1 canvas bag is 4 yard canvas 1 yard leather

So in order to make 12 leather bags u would need 36 yards of Leather which leaves u 20 yards of leather and 24 yards of canvas to make 20 canvas bags

and then to make 20 canvas bags u would need 80 yards of canvas which leaves u with 0 yards of canvas left and u need 20 yards of leather which is left over so all materials are used

So how do i implement that table into your equation?