Expert: Paul Klarreich - 10/3/2007

Question
Hi Paul,

I'm a student currently studying mathematical economics, a combination of maths and economics, which i found harder than just maths and i need your help with the question below:

The revenue (in dollars per square metre) obtained from the sale of tomatoes grown in a heated greenhouse is given by:
R = 5T(1 - e^-x)

Where T is the temperature and x is the amount of fertilizer per square metre.
Fertilizer costs 20 dollars per unit. To heat a square metre to a certain temperature costs 0.1T^2 dollars.

(a) Derive the per square metre profit function for the tomato grower.

(b) Given that the grower would choose T and x to maximize profits, confirm that the extreme values of T and x are:
(T,t)=(20, ln(5)) and (T,t)=(5, ln(5/4))

(c) Use the second order conditions to determine whether either of these two points corresponds to a relative maximum or minimum

Thanks,
Camilla

Answer
Questioner:   camilla
Category:  Advanced Math
Private:  No
 
Subject:  Profit maximisation << spelled with a 'z' here.
Question:  Hi Paul,

I'm a student currently studying mathematical economics, a combination of maths and economics, which i found harder than just maths and i need your help with the question below:

The revenue (in dollars per square metre) obtained from the sale of tomatoes grown in a heated greenhouse is given by:
R = 5T(1 - e^-x)

Where T is the temperature and x is the amount of fertilizer per square metre.
Fertilizer costs 20 dollars per unit. To heat a square metre to a certain temperature costs 0.1T^2 dollars.

(a) Derive the per square metre profit function for the tomato grower.

(b) Given that the grower would choose T and x to maximize profits, confirm that the extreme values of T and x are:
(T,t)=(20, ln(5)) and (T,t)=(5, ln(5/4))

(c) Use the second order conditions to determine whether either of these two points corresponds to a relative maximum or minimum

Thanks,
Camilla
.......................................
Hi, Camilla,

I assume that the profit = revenue - expense

Revenue is given by your  R function, and expense seems to be:

cost of fertilizer = 20x
cost of heat = 0.1T^2

Revenue =  5T(1 - e^-x)
Cost = 20x + 0.1T^2

P =  5T(1 - e^-x) - (20x + 0.1T^2)
P =  5T(1 - e^-x) - 20x - 0.1T^2
........................................
So Profit is a function of two variables, x,T, so we will use PARTIAL DERIVATIVES, normally a part of third-semester calculus, which I'm sure you must have completed already if they let you into business school.
........................................
For your 'optimization', we need:

dP/dx = 0   and  dP/dt = 0

dP/dx = 5T(e^-x) - 20

dP/dt = 5(1 - e^-x) - 0.2T

Set them equal to zero and you have a pair of simultaneous equations.

5T(e^-x) - 20 = 0       (A)
  
5(1 - e^-x) - 0.2T = 0  (B)

T = 4e^x, after solving A for T.

Call  e^x = E, just for ease in typing.  Now  T = 4E.

5(1 - 1/E) - 0.8E = 0

5 - 5/E = 0.8E

5E - 5 = 0.8E^2

25E - 25 = 4E^2

4E^2 - 25E + 25 = 0

Factor:

(4E - 5)(E - 5) = 0

Solutions are:

E = 5/4,  so  e^x = 5/4,  x = ln(5/4), and
T = 4E = 5

AND

E = 5,  so  e^x = 5,  x = ln(5)
T = 4E = 20.

There's your solutions,  (5,ln(5/4), and (20,ln(5))
..................................
Now for your tests:  There is a 'Second Derivative Test' which is described in nice mathematical notation at:

http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node70.html

(and probably a few other places)
So you compute the second derivatives:

We had:  dP/dx = 5T(e^-x) - 20  and we compute:

ddP/dxx = -5Te^-x
ddP/dxT = 5e^-x

We had:  dP/dt = 5(1 - e^-x) - 0.2T   and we compute:

ddP/dtt = -0.2
ddP/dTx = 5e^-x,  same as  ddP/dxT, of course.

Now compute: (shortened notation)

D = Pxx Ptt - Pxt Ptx

at each of your points, and apply the rule you will see at that URL.  I'll leave that part to you.