FINDING EQUATION - HELP

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November 4th 2008, 08:58 AM
cnmath16

FINDING EQUATION - HELP

The Question ...

A clothing manufacturer has 100m of silk and 180m of wool. To make a suit requires 2m of silk and 3m of wool, and to make a dress requires 1m of silk and 2m of wool. If the profit on a suit is $108 and the profit on a dress is $60, how many suits and dresses should the manufacturer make to maximize the profit?

What I have so far...

Let x represent silk
Let y represent wool
Profit(suit)= $108
Profit(dress)= $60
Equation for suit: 2x + 3y = 108 ??
Equation for dress: 1x + 2y = 60 ??
Im not sure if these are right or not. Any help would be greatly appreciated!!

I am really just looking for the overall equation but any other tips would also make things easier! this is due tomorrow ahhh

Thanks in advance!
November 4th 2008, 10:11 AM
earboth

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Quote:

Originally Posted by cnmath16

The Question ...

A clothing manufacturer has 100m of silk and 180m of wool. To make a suit requires 2m of silk and 3m of wool, and to make a dress requires 1m of silk and 2m of wool. If the profit on a suit is $108 and the profit on a dress is $60, how many suits and dresses should the manufacturer make to maximize the profit?

What I have so far...

Let x represent silk
Let y represent wool
Profit(suit)= $108
Profit(dress)= $60
Equation for suit: 2x + 3y = 108 ??
Equation for dress: 1x + 2y = 60 ??
Im not sure if these are right or not. Any help would be greatly appreciated!!

I am really just looking for the overall equation but any other tips would also make things easier! this is due tomorrow ahhh

Thanks in advance!

Let x denote the number of suits and y the number of dresses.

Then the "consumption" of silk is described by: $2x+y\leq100$ and

Then the "consumption" of wool is described by: $3x+2y\leq180$

With the additional conditions $x\geq 0~\wedge~y\geq 0$ you'll get a pentagon within all points which satisfy these conditions.

The profit is calculated by:

$p= 108x+60y~\implies~y=-\dfrac95 x + \dfrac1{60}p$

Now choose a point of the pentagon such that the y-intercept (which contains the profit) is as large as possible.

Some results for confirmation: (20, 60),
$\dfrac1{60} p = 96$
November 4th 2008, 02:12 PM
cnmath16

Thanks a lot for your quick response! Are you saying that the answer to the question is that approximately 96 suits and dresses should be made from the manufacturer in order to maximize the profit? Or what does the point you chose (20,60) represent?
November 4th 2008, 10:22 PM
earboth

Quote:

Originally Posted by cnmath16

Thanks a lot for your quick response! Are you saying that the answer to the question is that approximately 96 suits and dresses should be made from the manufacturer in order to maximize the profit? No

Or what does the point you chose (20,60) represent?

If the straight line passes through the point (20, 60) the y-intercept gets its largest value.

The coordinates indicate that the manufacturer should produce

x = suits = 20 and
y = dresses = 60

then the profit can be calculated by

$<br /> \dfrac1{60} p = 96~\implies~p=96 \cdot 60 = 5760<br />$

This is the maximum profit.