Optimization word problem: demands given by QEU= 1/5(105-PEU) and QUS= 1/6(71-PUS)

[belgiumrob]

Hi everyone!

So I got this optimization word problem, and I'm kind of stuck, really hoping that someone could point me out to the right direction?

The problem: "A European winegrower wants to sell champagne in the USA as well. The demand for this champagne in the EU market and the US market are denoted by: QEU= 1/5(105-PEU) and QUS= 1/6(71-PUS). The total cost is given by TC = q^2+15q+20. How many bottles should be brought on respectively the EU market and the US market?"

So, first i changed the demand fuctions, to get a function on price:
PEU= 105-5QEU
PUS= 71-6QUS

Then i thought q = QUS+QEU, so:
q= 1/5(105-PEU) + 1/6(71-PUS) or q = (21-1/5PEU) + (71/6 - 1/6PUS)

Then I thought Total Revenue would be:
TR= (105-5QEU)*QEU + (71-6QUS)*QUS

And thus Total profit would be TR-TC, so:
(105-5QEU)*QEU + (71-6QUS)*QUS - ((21-1/5PEU) + (71/6 - 1/6PUS))^2 + 15((21-1/5PEU) + (71/6 - 1/6PUS)) + 20


And now I'm stuck, where do i go from here?

Thanks for your replies...

 

[belgiumrob]

Hi everyone,

IS there no one with a clue of how to solve this problem further?

 

[Ishuda]

Hi everyone,

IS there no one with a clue of how to solve this problem further?

Formulate the problem in terms of QUE and QUS [actually, just to make writing easier, I would replace QUE by x and QUS by y], i.e. replace (21-1/5PEU) + (71/6 - 1/6PUS) by QUE+QUS. You will then have a formula in two variables. To solve the problem you then need to find critical points of a function of two variables. An example of this type problem can be found at
http://math.oregonstate.edu/home/pr...usQuestStudyGuides/vcalc/min_max/min_max.html


EDIT: Just to re-write the profit (P) equation
P(x,y) = (105 - 5 x) x + (71 - 6 y) y - [(x + y)^2 + 15 (x + y) + 20]
P(x,y) = 105 x - 5 x² + 71 y - 6 y² -[x² + 2 x y + y² + 15x + 15 y + 20]
P(x,y) = - 6 x² - 7 y² - 2 x y + 90 x + 56 y - 20

Highlight below for a solution
\(\displaystyle \frac{\partial\, P(x,y)}{\partial\, x}\, =\, -12 x\, -\, 2\, y\, +\, 90\)
\(\displaystyle \frac{\partial\, P(x,y)}{\partial\, y}\, =\, -14 y\, -\, 2\, x\, +\, 56\)
For critical points both of these must be zero, so
y = (-6 x + 45)
and
-14 (-6x + 45) - 2 x + 56 = 82 x - 574 = 0
or x = 7 and y = 3.

Now
\(\displaystyle \frac{\partial^2\, P(x,y)}{\partial\, x^2}\, =\, -12\)
\(\displaystyle \frac{\partial^2\, P(x,y)}{\partial\, y^2}\, =\, -14\)
\(\displaystyle \frac{\partial^2\, P(x,y)}{\partial x\, \partial y}\, =\, -\, 2\)
so
D(x,y) = \(\displaystyle \frac{\partial^2\, P(x,y)}{\partial\, x^2}\, \frac{\partial^2\, P(x,y)}{\partial\, y^2}\, -\, [\frac{\partial^2\, P(x,y)}{\partial x\, \partial y}]^2\, =\, 164\)
and D(7,3) > 0 meaning P(7,3) is a relative maximum.