Linear Projection

I've been out of High School for twenty years now. When I was in school, I was pretty good in math, but I've lost a lot of the higher stuff I learned through lack of use. Still, up till now, all I ever needed to figure out my daughters' homework was a quick google search or some light reading in their textbook. Now, though, my second daughter is doing things in her 9th grade "Advanced Algebra" class that I can barely grasp the workings of.

The question is below:

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Linear projection.
Question says this company makes chairs and tables. Tables take 3 hours of carpentry and 2 hours of finishing and make a 35 dollar profit. Chairs take 2 hours of carpentry and 1/2 an hour of finishing and make a 20 dollar profit. They can use up to 108 carpentry hours and 20 finishing hours daily. How many chairs and tables should be made each day to maximize profit?

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Now I need help... because she doesn't have a clue how to do it, and I only have the vaguest notions. I need some pointers on figuring out just how to solve this problem so that I can give her the tools to solve it. From googling and looking up "Linear Projection" online, I think I'm supposed to generate some equations for graphing, and then find the intersections, one of which will be the point at which so many tables and so many chairs equal the greatest profit. Where I'm stuck is that I can't seem to figure out just which variables to use and what the equations are I need to be solving!

My first thought was that I needed one equation for tables and another for chairs, with X and Y representing hours of Carpentry and Finishing respectively:

3x + 2y = 35
2x + .5y = 20

I solved these equations for y:

y = (35 - 3x) / 2
y = 40 - 4x

And then I get the feeling I'm heading in the wrong direction.. especially when I start graphing them and things seem to be going in the wrong direction.


Anyway, at this point any help would be vastly appreciated, as I'm thoroughly lost on my own.

Hey RugerWulf.

For this problem you should consider writing your profit equations in terms of hours where you have h1 + h2 = 108 and h3 + h4 = 20 with the equation for the profits being P1 = 35*(3*h1 + 2*h3) and P2 = 20*(2*h2 + 0.5*h4) where total profit is P = P1 + P2 where "profit" is in terms of profit-hours not just profit (different units).

So now you have to maximize P given the constraints. You can get things in terms of just h1 and h3 using the substitution.

Now you have a bivariate function and you can differentiate to find the minimum and maximums of the function taking into account the conditions for the hours.

The standard way to do optimization problems of this type is to use what is known as Lagrange multipliers: have you used them before?

Quote:

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Originally Posted by chiro

Hey RugerWulf.

For this problem you should consider writing your profit equations in terms of hours where you have h1 + h2 = 108 and h3 + h4 = 20 with the equation for the profits being P1 = 35*(3*h1 + 2*h3) and P2 = 20*(2*h2 + 0.5*h4) where total profit is P = P1 + P2 where "profit" is in terms of profit-hours not just profit (different units).

So now you have to maximize P given the constraints. You can get things in terms of just h1 and h3 using the substitution.

Now you have a bivariate function and you can differentiate to find the minimum and maximums of the function taking into account the conditions for the hours.

The standard way to do optimization problems of this type is to use what is known as Lagrange multipliers: have you used them before?

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I'm afraid you lost me at bivariate.

From what it looks like you're saying, you're measuring Profit in terms of the number of hours taken to produce each item.. Where the question is asking for the number of items needed to produce the most profit in $. Either I'm getting more confused or you're going in another direction entirely.

As for Lagrange multipliers, I'll look them up, but I don't recall hearing of them before.

Bi-variate just means two independent variables like z = f(x,y) instead of z = f(x).

The starting point for doing optimization problems with constraints is with Lagrange Multiplier techniques.

i think you're doing it right, but you're solving for the wrong variables.

let x be the number of tables made, and y the number of chairs.

the daily profit is: 35x + 20y <---this is what we want to maximize. but we have certain constraints:

if we make x tables and y chairs, we use up 3x + 2y hours of carpentry. since we can only use up 108, we must have: 3x + 2y ≤ 108.

similarly, we use up 2x + y/2 hours of finishing each day, so we must have: 2x + y/2 ≤ 20.

if you graph y = 54 - (3/2)x and y = 40 - 4x, you can see that the controlling factor is the finishing hours (makes sense: we have fewer of these to allocate).

that is: our "feasible region" is the area under the line y = 40 - 4x, and bounded by the x-axis and y-axis (we cannot produce "negative" chairs or tables).

the line of greatest increase in profits is the gradient: y = (7/4)x. the optimum solution lies on the intersection of a line perpendicular to the gradient, which intersects our feasible region with the greatest y-intercept.

such a line is of the form y = -(4/7)x + c (if you're wondering the 7 and the 4 come from the ratio of tables to chairs). now we need to solve for c. but this is easy: the maximum value for c occurs when x = 0,

and at this point the only choice we have for y is y = 40. so the best strategy for the company is to make 40 chairs every day, for $800 profit.

let's do a sanity check. suppose the company make 1 table. this uses up 3 hours of carpentry and 2 hours of finishing. now they have enough carpentry hours to make 52 chairs, but only enough finishing hours to make 36 chairs.

1*35 + 36*20 = $755, which is $45 less.

if they make 2 tables, they can make 51 chairs with the carpentry hours, but only 32 chairs in all, because of the finishing. so now their profit is:

2*35 + 32*20 = $710, $45 less than making just 1 table.

continuing this way, we see that each table we make cost the company $45 in potential profits. the maximum number of tables we might make is 10 (we could make more with carpentry hours, but we couldn't finish them).

if i owned that company, i'd hire some more finishers or lay off some carpenters, because even only making chairs, we have 28 unused hours of carpentry a day.

Deveno: is this like a simplex problem?