I’ve just finished my textbook, “Forgotten Calculus”. The last chapter is on multivariable calculus. Disappointingly, while there were some “pure math” problems on finding and classifying relative extrema, there were no real-world applications of this technique, i.e optimization problems.

So I decided to make one up, or rather base it on a marginal profit problem in the text. So what I’ve done below is not an “assigned” question per se. So the book has no answer for me to check myself by.

I would therefore ask any of you kindly, mentoring types out there to tell me if I am going about this right or wrong (I suspect the latter) and if need be correct me. To make it easier I show pretty much all of my work, although it does make this post pretty long.

I also need some guidance on interpretation.

Lastly, if anyone can recommend any sources which have well-explained, worked examples in multivariate optimization, I’d be grateful. I don't think I need (or have time for) a whole book on the subject; 5-10 examples would help a lot.

The premise

Total annual profit z from a small “summer hobby” farm is determined as

z = f(x,y) = 10x^3 + 20y^2 – 10xy

when x = weekly ad spending = 10 and

y = acres farmed = 5.

This much came from the text.

The question and strategy summary

I’d like to find out which combination(s), if any, of ad spending and acres maximizes (or minimizes) profit.

I will attempt to do so by trying to locate and evaluate the critical points, and hope they make some sense.

By the way: as a reference point, I evaluate the original function at values given, i.e

If

f(x,y) = 10x^3 + 20y^2 – 10xy

then

f(10, 5) = 10 (10)^3 + 20(5)^2 – 10 (10) (5)

= 10,000 + 500 - 500

=10,000

Find critical points

Start by taking the partials f[subscript]x and f[subscript]y

f[subscript]x = 30x^2 – 10y

f[subscript]y = 40y – 10x

Next, solve for x and y by setting each partial equal to 0, and solving as a “system of equations”, if this is the right term.

0 = 30x^2 – 10y

0 = 40y – 10x

Rearrange the bottom equation…

0= 30x^2 - 10y

0= -10x + 40 y

Put them into a common format…

0 = 30x^2 – 0x – 10y

0 = 0x^2 – 10x + 40y

Multiply the top equation by 4 to get the “y”s to cancel out when I then “add” the two equations together….

0 = 120x^2 – 0x – 40y

0 = 0x^2 – 10x + 40y

___________________

0 = 120x^2 – 10x +0

Use the quadratic formula (done on a calculator, work not shown) to solve for x

Point 1: x = .08333(displayed in this explanation showing only 3 threes, but in calculations, using 6 threes)

Point 2: x = 0

Solve for y at Point 1 by plugging the Point 1 x value into one of the first partials set equal to 0

f[subscript]x = 30x^2 – 10y

0 = 30x^2 – 10y

0 = 30(.08333)^2 – 10y

0 = .208333 – 10 y

.208333 = 10y

Point 1: y = 48

So Point 1 = (.08333; 48; z)

Solve for y at point 2 using the same method, only using the Point 2 x value

f[subscript]x = 30x^2 – 10y

0 = 30x^2 – 10y

0 = 30 (0) – 10y

0 = 10y

Point 2: y = 0

So Point 2 = (0; 0; z)

Find z for Point 1

Plug the Point 1 x and y values into the original function

z = 10x^3 + 20y^2 – 10xy

= 10 (.08333)^3 + 20(48)^2 - 10 (.08333)(48)

= .005787 + 46,080 – 39.999

Point 1: z = 46,040

So Point 1 = (.08333; 48; 46,040)

Find z for Point 2 using the same method, only plugging in Point 2 x values

z = 10x^3 + 20y^2 – 10xy

= 0 + 0 -0

Point 2: z = 0

So Point 2 = (0; 0; 0)

An aside

At this point, before properly classifying the critical points, I was thinking ahead: “Ok, this makes sense – Point 1 is going to be a maximum, at which the maximum profit is 46,060, which is within the same order of magnitude as the profit under the premise conditions, and therefore seems plausible; while Point 2, showing 0 profit when there is no ad spending and no acres to far, makes perfect sense as a minimum”

As you’ll see below, I was wrong…

Classifying the critical points

Let D (x,y) = [ ( f[subscript]xx ) ( f[subscript]yy ) ] - (f[subscript]xy )^2

From above, for reference:

f[subscript]x = 30x^2 – 10y

f[subscript]y = 40y – 10x

So

f[subscript]xx = 60x

f[subscript]yy = 40

f[subscript]xy = 10

So

D= [ (60x) (40) ] - (10)^2

D= 240x – 100

For Point 1:

D = 240(.08333) - 100

D = 20 – 100

D = - 80

This is negative; thus Point 1 is a saddle point

For Point 2:

D = 240(0) - 100

D = – 100

This is negative; thus Point 2 is also a saddle point

Concluding questions

First: Have I conceived and / or executed the solution wrong?

Second: I must say this is somewhat unsatisfying. Assuming I’ve done this right, I’m not sure what to make of the answer. I know what a drawing of a saddle looks like but frankly don’t understand what it means in this case – a profitability problem with no minima or maxima, only two saddle points. How would you plan an optimal business strategy based on this?

If you’ve read this far, thanks for your attention.

Regards,

lingyai