# Business optimization eek!

November 22, 2009 3:55 AM Subscribe

I've got these 2 math problems that are driving me crazy. Business optimization. Please help.

I just can't grok this concept! Help me figure it out?

Q#1 A business consultant advises a client that by spending extra on advertising, the client will increase profits. For an increase of $x in advertising, profit will increase by 20 x

Q#2 Fred has a business that makes two kinds of dining room suites. Model A takes 4 hours to assemble and 2 hours to finish. Model B takes 3 hours to assemble and 4 hours to finish. 3 employees assemble the suites and 2 do the finishing work. All employees work a 36 hour week.

Profit on Model A is $200, on Model B, $260. A minimum of 4 of each model has to be manufactured each week.

What production of each model will maximize profit? What is the maximum profit?

I just can't grok this concept! Help me figure it out?

Q#1 A business consultant advises a client that by spending extra on advertising, the client will increase profits. For an increase of $x in advertising, profit will increase by 20 x

^{2}e

^{-.5x}. What increase in expenditure will yield the greatest increase in profit? (x is the amount in 100s of dollars.)

Q#2 Fred has a business that makes two kinds of dining room suites. Model A takes 4 hours to assemble and 2 hours to finish. Model B takes 3 hours to assemble and 4 hours to finish. 3 employees assemble the suites and 2 do the finishing work. All employees work a 36 hour week.

Profit on Model A is $200, on Model B, $260. A minimum of 4 of each model has to be manufactured each week.

What production of each model will maximize profit? What is the maximum profit?

do you have the 'solver' add-in on excel? for if you don't want to bother doing these by hand

posted by moorooka at 5:13 AM on November 22, 2009

posted by moorooka at 5:13 AM on November 22, 2009

Or graph the first one on a graphing calculator (or one of many programs on the web which will do so.)

posted by Obscure Reference at 5:21 AM on November 22, 2009

posted by Obscure Reference at 5:21 AM on November 22, 2009

If you are doing these for a course and dont understand them you need some extra math help. If you are doing them for some other reason and just need the answer you can try typing them into wolframalpha.

posted by shothotbot at 6:11 AM on November 22, 2009

posted by shothotbot at 6:11 AM on November 22, 2009

Here's a graph of your exponential function, you can see it peaks around 4. You can see on the plot of the derivative that it actually crosses zero at exactly 4. In calculus, the derivative is "the amount of change" at each point. So $400 is exactly how much you need to increase your spending and you should see an increase in profit of 43.3073*100 = $433.74

If you think about a pendulum swinging, it swings the fastest when it's perpendicular to the ground, and it gets slower and slower until it actually changes direction and starts going the other way. The derivative of of the pendulum's

posted by delmoi at 6:29 AM on November 22, 2009

If you think about a pendulum swinging, it swings the fastest when it's perpendicular to the ground, and it gets slower and slower until it actually changes direction and starts going the other way. The derivative of of the pendulum's

*location*is it's*speed*. You can think about your function the same way. The "speed" of the change is zero at 4, which the point where it's actually changing direction. In math, this is called the "inflection point"_{The odd thing is, if you go negative, and reduce the spending, the profit goes up exponentially, by reducing expenses by $500, you would save $296,826,000. I assume that's just a mistake in the way the question was expressed, and they intended to say that you could only increase by a positive number (no 'increase of -10' allowed)}posted by delmoi at 6:29 AM on November 22, 2009

_{(oops, that's reducing expenses $1000 in the 3rd paragraph above)}

posted by delmoi at 6:48 AM on November 22, 2009

Yeah, delmoi's got it. If you want to do it analytically:

profit:

P(x) = 20 * x^2 * exp(-x/2)

At the maximum, the derivative of P will be zero:

dP/dx = 0

dP/dx = 40 * x * exp(-x/2) - 10 * x^2 * exp(-x / 2) = 0

4 * x = x^2

x = 4

(the derivative is also zero at x = 0, but that's a minimum)

So x = $400 will give you a maximum increase in profit.

As for the other one, I don't have time to work it out, but start by figuring out what quantities you can vary, and what are fixed, and make a table to sort it all out.

I.e. the quantities you can vary are the number of suite A and B that you can make each week. Call that Na and Nb. Figure out what the constraints are on them. How is the profit related to Na and Nb? It's a little trickier since you can't make fractions of a suite, so you'll have to round the right way to maximize profit.

posted by Salvor Hardin at 7:15 AM on November 22, 2009

profit:

P(x) = 20 * x^2 * exp(-x/2)

At the maximum, the derivative of P will be zero:

dP/dx = 0

dP/dx = 40 * x * exp(-x/2) - 10 * x^2 * exp(-x / 2) = 0

4 * x = x^2

x = 4

(the derivative is also zero at x = 0, but that's a minimum)

So x = $400 will give you a maximum increase in profit.

As for the other one, I don't have time to work it out, but start by figuring out what quantities you can vary, and what are fixed, and make a table to sort it all out.

I.e. the quantities you can vary are the number of suite A and B that you can make each week. Call that Na and Nb. Figure out what the constraints are on them. How is the profit related to Na and Nb? It's a little trickier since you can't make fractions of a suite, so you'll have to round the right way to maximize profit.

posted by Salvor Hardin at 7:15 AM on November 22, 2009

delmoi- I was considering posting that, but I decided we could make rational assumptions.

For question #2, you need to make some assumptions. The assumption I would make is that the company employs exactly 3 assemblers and 2 finishers, and that people are dedicated to that task. If you don't make those assumptions, the question is ill-formed. If you DO make that assumption, then the number of assemblers and finishers is not relevant information. Although this sounds like a homework question, I will show you how to set it up, without solving it for you.

X = # of hours of the assembly team

Y = # of hours of the finishing team

A = # of model A made

B = # of model B made

Profit = A * $200 + B * $260

X = 4 * A + 3 * B

Y = 2 * A + 4 * B

Constraints:

X <= 36

Y <= 36

A >= 4

B >= 4

Given these contraints, maximize Profit.

posted by JMOZ at 7:19 AM on November 22, 2009

For question #2, you need to make some assumptions. The assumption I would make is that the company employs exactly 3 assemblers and 2 finishers, and that people are dedicated to that task. If you don't make those assumptions, the question is ill-formed. If you DO make that assumption, then the number of assemblers and finishers is not relevant information. Although this sounds like a homework question, I will show you how to set it up, without solving it for you.

X = # of hours of the assembly team

Y = # of hours of the finishing team

A = # of model A made

B = # of model B made

Profit = A * $200 + B * $260

X = 4 * A + 3 * B

Y = 2 * A + 4 * B

Constraints:

X <= 36

Y <= 36

A >= 4

B >= 4

Given these contraints, maximize Profit.

posted by JMOZ at 7:19 AM on November 22, 2009

I will give the one additional hint that there are only 6 possible combination (A, B) that meet the constraints. I am, of course, assuming that A and B must be integers, and there is no work carried over into the next week.

posted by JMOZ at 7:26 AM on November 22, 2009

posted by JMOZ at 7:26 AM on November 22, 2009

JMOZ, aren't the constraints X <= 108 (3 assemblers working 36 hours apiece) and Y <= 72 (2 finishers working 36 hours apiece)?

posted by cali59 at 7:37 AM on November 22, 2009

posted by cali59 at 7:37 AM on November 22, 2009

Cali59- No, X is the hours of the assembly DEPARTMENT. Given the parameters of the question, the fact that department has 3 people is irrelevant. I assume 2 assembler can't make a set alone in triple the time.

posted by JMOZ at 7:58 AM on November 22, 2009

posted by JMOZ at 7:58 AM on November 22, 2009

Usually these questions are phrased "If you increase expenditure by x dollars, then

P(x) = R + f(x) - (E + x)

to maximize P(x) set P'(x) = 0.

P'(x) = f'(x) -1.

It is a slight difference from the problem you've given. Hopefully this helps rather than add to the confusion.

posted by water bear at 9:15 AM on November 22, 2009

*revenue*increases by f(x) dollars. Find the increase in expenditure that maximizes profit." Profit = Revenue - Expenditure.P(x) = R + f(x) - (E + x)

to maximize P(x) set P'(x) = 0.

P'(x) = f'(x) -1.

It is a slight difference from the problem you've given. Hopefully this helps rather than add to the confusion.

posted by water bear at 9:15 AM on November 22, 2009

While I'm certainly no math whiz and won't attempt these equations, the logical business person in me with several years experience working at ad agencies and doing marketing in-house have me questioning the first question.

What is the medium they would increase your spend in? Is it primarily for branding or direct response (like AdWords or direct mail)? If it is for branding, then I can understand the complexity of the equation since the impact of branding truly becomes felt when you expand your reach even further and gain critical mass. However for direct response, you have a conversion rate. Once you have a normalized conversion rate, in theory every extra dollar spent converts at the same rate and thus your profits scale proportionately with your spend.

I also have to question either the equation you've given or Delmoi's method of solving it (no offense). Again, not a math whiz here but to debate over whether to spend an extra $400 to see an extra $433.74 in profit seems a bit off if that math doesn't scale. If that is the maximum you could increase profits by (as a result of increasing your ad spend) then you've most likely already blown your profits just by having a single billable meeting with this "consultant."

posted by Elminster24 at 10:08 AM on November 22, 2009

What is the medium they would increase your spend in? Is it primarily for branding or direct response (like AdWords or direct mail)? If it is for branding, then I can understand the complexity of the equation since the impact of branding truly becomes felt when you expand your reach even further and gain critical mass. However for direct response, you have a conversion rate. Once you have a normalized conversion rate, in theory every extra dollar spent converts at the same rate and thus your profits scale proportionately with your spend.

I also have to question either the equation you've given or Delmoi's method of solving it (no offense). Again, not a math whiz here but to debate over whether to spend an extra $400 to see an extra $433.74 in profit seems a bit off if that math doesn't scale. If that is the maximum you could increase profits by (as a result of increasing your ad spend) then you've most likely already blown your profits just by having a single billable meeting with this "consultant."

posted by Elminster24 at 10:08 AM on November 22, 2009

Elminster- are you really surprised that the questions in what is effectively a match class are contrived?

Optimization is a basic principle of business/engineering, and using contrived questions is better from a pedagogical point of view because it allows well-posed questions to be asked. If profits increase without bounds with increased advertising spending, then the optimum ad spending is, of course, infinite. This, clearly, isn't realistic, either. Of course, it would be better to ask questions where the model isn't entirely made-up, but it would also be good if the questions were clearer, as you can see from the different interpretations that cali59 and I used. Also note that his/her interpretation is just as valid as mine and also gives a good answer, albeit a different one.

posted by JMOZ at 10:33 AM on November 22, 2009

Optimization is a basic principle of business/engineering, and using contrived questions is better from a pedagogical point of view because it allows well-posed questions to be asked. If profits increase without bounds with increased advertising spending, then the optimum ad spending is, of course, infinite. This, clearly, isn't realistic, either. Of course, it would be better to ask questions where the model isn't entirely made-up, but it would also be good if the questions were clearer, as you can see from the different interpretations that cali59 and I used. Also note that his/her interpretation is just as valid as mine and also gives a good answer, albeit a different one.

posted by JMOZ at 10:33 AM on November 22, 2009

JMOZ, your response prompted me to reread the question and it appears I completely took it the wrong way. I must have read it too quickly and thought this guys company hired a consultant that was telling them these things and he was looking for help in justifying it. TOTALLY missed that this was just a basic text-book math problem for school. Oops.

In real life there are a lot more variables to consider as I stated in my post though and it is important to keep the "unplannables" in mind. And yes, optimization is a huge part of things.

posted by Elminster24 at 3:25 PM on November 22, 2009

In real life there are a lot more variables to consider as I stated in my post though and it is important to keep the "unplannables" in mind. And yes, optimization is a huge part of things.

posted by Elminster24 at 3:25 PM on November 22, 2009

This thread is closed to new comments.

2) looks like a pair of simultaneous equations to me. Should be easy to solve.

posted by pharm at 4:26 AM on November 22, 2009