Maximizing Profit with x Units: A Math Puzzle

[Yankel]

Hello,

I need some help with this question, it is not very clear and I find it hard to translate from text to math...

Some company has found that x units of it's product can be sold for p$ a unit when x=1000-p. The cost of producing x units in a day is C(X)=3000+20x.

A. Found the income function R(X) and the profit function P(X).

B. Assuming that the company can produce 500 units a day, determine how many units they need to produce and sell for maximal profit.

C. What is the maximal profit ? What is the price of a unit that leads to a maximal profit ?

Thanks...

 

[MarkFL]

A.) Revenue (income) in dollars is price per unit times units sold:

$\displaystyle R(x)=x(1000-x)=1000x-x^2$

Profit is revenue less costs:

$\displaystyle P(x)=R(x)-C(x)=(1000x-x^2)-(3000+20x)=-x^2+980x-3000$

B.) Since this is posted in the calculus forum, now set:

$\displaystyle P'(x)=0$

to maximize the profit. Is $\displaystyle 0\le x\le500$?

i) How can you do this without calculus?

ii) How do you know you have a maximum?

C.) What do you think you need to do, given the above result, to find the maximum profit and the price per unit that maximizes profit?

 

[Yankel]

right, I have calculated P'(x)=0 and got x=490, it is between 0 and 500. don't i have to check the edges, i.e. x=500 to see if it's a local maximum or global ?
x=490 is maximum, I used P''(x) to check it.

I am still not sure about part C.

 

[MarkFL]

Yes, you have the correct critical value of x = 490, and using the second derivative is exactly what I had in mind, since it is a negative constant, we know the curve is concave down for all x. Thus, we need not check the end-points as we know the extremum we found is a global maximum.

To do this without calculus, we see that the profit is parabolic, and opens downward. Thus, we need only find the axis of symmetry to find the critical value:

$\displaystyle x=-\frac{980}{2(-1)}=490$

Now, for part C, since we know $\displaystyle x=490$ yields the maximum profit, to find the maximum profit means we need to evaluate:

$\displaystyle P(490)$

To find the price at which $\displaystyle x=490$, we need to use the given relationship between the price and the number of units:

$\displaystyle x=1000-p\,\therefore\,p(x)=1000-x$

Thus, we need to evaluate:

$\displaystyle p(490)$

What do you find?

 

[Yankel]

Ok, so

P(490) = 237,100

but the unit price, did you mean 510 ?

 

[MarkFL]

Yes for both:

$\displaystyle P(490)=-(490)^2+980(490)-3000=237100$

$\displaystyle p(490)=1000-490=510$

Good work!

 

[HallsofIvy]

MarkFL also asked "How can you do this without calculus?"

In this problem the profit function is -x²+ 980x- 3000, a quadratic function. Its graph is a parabola opening downward and so it maximum value is at its vertex. And you can find that by completing the square.