Maximising Profit: Solving Linear Demand Functions

[sand7000]

I have been trying to figure this out all day. please help.

If the cost of one unit is $10, 20 units will be sold. If the cost is $11, 18 units will be sold. a) Find the demand function assuming it is linear.
b) If the materials for each unit cost $6 what should the selling price be to maximise profit?

I believe that the demand function is p(x)=-.5x+20 but I cannot figure out how to maximise profit. PLEASE HELP!

 

[Chen]

If d(p) is the demand as a function of the price, you know that:

d(p) = Ap + B

You need to find A and B. You have two pairs of values for d(p) and p, so use them to find A and B:

20 = 10A + B
18 = 11A + B

The function you posted does not seem correct.

The cost of all units would be 6*d(p). The return value would be d(p)*p. The profit is the difference:

f(p) = d(p)*p - 6*d(p)

Now replace d(p) with the actual function:

f(p) = (Ap + B)*p - 6*(Ap + B) = Ap² + Bp - 6Ap - 6B

Find the derivative of f(p), see where it equals zero, etc etc.

 

[M98Ranger]

a) To find the demand function, we can use the two given points (10,20) and (11,18) and use the slope-intercept form of a linear function, y=mx+b, where m is the slope and b is the y-intercept.
First, we can find the slope, m, by using the formula (y2-y1)/(x2-x1), where (x1,y1) is one point and (x2,y2) is the other point. In this case, (x1,y1) = (10,20) and (x2,y2) = (11,18). So, m=(18-20)/(11-10)=-2/1=-2.
Next, we can plug in one of the points into the equation and solve for b. Let's use (10,20). So, 20=-2(10)+b. Solving for b, we get b=40.
Therefore, the demand function is p(x)=-2x+40.

b) To maximize profit, we need to find the selling price that will result in the highest revenue, which is the product of the selling price and the number of units sold. In this case, the selling price is p(x) and the number of units sold is x.
So, the revenue function is R(x)=xp(x)=-2x^2+40x.
To find the maximum revenue, we can use the vertex formula x=-b/2a, where a=-2 and b=40.
So, x=-40/(2(-2))=10.
Therefore, to maximize profit, the selling price should be p(10)=-2(10)+40=$20 per unit.