A retailer has been selling 1200 tablet computers a week at $350 each. The marketing department estimates that an additional 80 tablets will sell each week for every $10 that the price is lowered.
a) Find the demand function.
b) What should the price be set at in order to maximize revenue?
c) If the retailer's weekly cost function is
C(x) = 35,000 + 120x
what price should it choose in order to maximize its profit?
2 Answers

For every $10 discount, 80 more are sold.
So if the original price of $350 sold 1200 per week, then if we set "x" to be the number of $10 discounts, then the new price becomes:
(350  10x)
And the number of units sold will be:
(1200 + 80x)
the revenue is the product of the cost per unit and the number of units sold, so we can solve for that in terms of x:
R(x) = (350  10x)(1200 + 80x)
R(x) = 420000 + 28000x  12000x  800x²
R(x) = 420000 + 16000x  800x²
To find the maximum revenue, we can solve for the zero of the first derivative to get the number of $10 discounts to apply:
R'(x) = 16000  1600x
0 = 16000  1600x
1600x = 16000
x = 10
So if ten $10 discounts are to be applied to get the maximum revenue, then the cost becomes:
350  10 * 10
350  100
$250 per unit
Let's determine how many units will be sold at this price:
1200 + 80x
1200 + 80 * 10
1200 + 800
2000 units

We are then given the cost function:
C(x) = 35000 + 120x
Presuming that is in terms of units produced, so before we can find the profit function, we need to rework our revenue function to also be in terms of units sold (our first one was in terms of $10 discounts).
Let's start with the vertex form of a quadratic:
R(x) = a(x  h)² + k
Where the vertex is the point (h, k). We know this as we solved for the price and units sold in part 1. At a price of $250 each, you sell 2000 units, which is a total revenue of $500,000
So the vertex is the point (2000, 500000). That is our h and k.
We also have another point given the original baseline of 1200 units sold at $350, which has a revenue of $420,000), so another data point on the curve is (1200, 420000) That is our x and y (or R(x)). We can substitute what we know and solve for a, then expand it to determine our quadratic:
420000 = a(1200  2000)² + 500000
80000 = a(800)²
80000 = 640000a
a = 80000 / 640000
a = 1/8
So our equation is:
R(x) = (1/8)(x  2000)² + 500000
Expand that out to get:
R(x) = (1/8)(x²  4000x + 4000000) + 500000
R(x) = (1/8)x² + 500x  500000 + 500000
R(x) = (1/8)x² + 500x
Now we have a revenue function in terms of units sold and a cost function in terms of units produced. Presuming you sell all that you manufacture, the profit is the difference between the two :
P(x) = R(x)  C(x)
Substitute what we know to come up with the profit function:
P(x) =(1/8)x² + 500x  (35000 + 120x)
and simplify:
P(x) =(1/8)x² + 500x  35000  120x
P(x) =(1/8)x² + 380x  35000
So what is the maximum profit? Do the same as above and solve for the zero of the first derivative:
P'(x) = (1/4)x + 380
0 = (1/4)x + 380
(1/4)x = 380
x = 380 * 4
x = 1520
To get the maximum profit, you would want to manufacture and sell 1,520 units.
We want the price for this level of sales, so let's go back to our original equations and find the number of $10 discounts that are needed to make this level of sales, then work back to find the cost at that level:
1200 + 80x = 1520
80x = 320
x = 4
So four $10 discounts are needed, making the price:
350  4 * 10
350  40
$310 per unit.

a) Find the demand function.
x1200 = the increase of units a week per $10 decrease
80 more units are sold per $10 decrease, therefore,
p(x) = 350(10/80)(x1200)
= 2001/8x <simplified
b) What should the price be set at in order to maximize revenue?
R(x) = xp(x)
=x(2001/8x)
=200x(1/8)x^2
R'(x)= 200(1/4)x
x=800 < R'(x)=0; solve for x
Maximized revenue at 800 more units sold
x1200=800
p(x)=350(10/80)(800) <substitute 800 for (x1200)
= 350100
=$250
c) If the retailer's weekly cost function is
C(x) = 35,000 + 120x
what price should it choose in order to maximize its profit?
P(x) = profit
P(x)= R(x)C(x)
= 200x(1/8)x^2(35000+120x)
= 200x(1/8)x^235000120x
= (1/8)x^280x+35000 < simplified and made x^2 term positive
P'(x)=(1/4)x80
x=240 < P'(x)=0; solve for x
Maximized profit as 240 units sold
x1200=240
p(x)=350(10/80)(240) <substitute 240 for x1200
= 35040
=$310