Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# help with calc!?

A manufacture has been selling 1050 television sets a week at \$420 each. A market survey indicates that for each \$26 rebate offered to a buyer, the number of sets sold will increase by 260 per week.

a) Find the function representing the demand 𝑝(𝑥), where 𝑥 is the number of the television sets sold per week and 𝑝(𝑥) is the corresponding price.

b)How large rebate should the company offer to a buyer, in order to maximize its revenue?

c) If the weekly cost function is 73500+140𝑥, how should it set the size of the rebate to maximize its profit?

Relevance
• 1 month ago

You have two data points for your demand function.

Using x is the number of units sold and y is the price, the points are:

(1050, 420) and (1310, 394)

1310 is 1050 + 260 and 394 is 420 - 26

We can find the equation for the line between those points:

y = mx + b

420 = m(1050) + b and 394 = m(1310) + b

420 - 1050m = b and 394 - 1310m = b

420 - 1050m = 394 - 1310m

260m = -26

m = -0.1

Now we can solve for b:

420 - 1050m = b

420 - 1050(-0.1) = b

420 + 105 = b

525 = b

The demand function is:

p(x) = -0.1x + 525

----

The total revenue is the product of the number of units sold (x) and the price per unit (p(x)):

R(x) = x p(x)

If we substitute the expression for p(x) into R(x) and simplify we get:

R(x) = x(-0.1x + 525)

R(x) = -0.1x² + 525x

The maximum revenue can be found by solving for the zero of the first derivative:

R'(x) = -0.2x + 525

0 = -0.2x + 525

0.2x = 525

x = 2625

Now we can solve for p(x) to get the price required to sell this many units:

p(x) = -0.1x + 525

p(x) = -0.1(2625) + 525

p(x) = -262.5 + 525

p(x) = 262.5

They have to sell at this price.  The rebate required is:

420 - 262.5 = \$167.50

-----

Now we are given the cost function:

C(x) = 140x + 73500

and want to know what the rebate needs to be to make a maximum profit.  Profit is the revenue minus the cost:

P(x) = R(x) - C(x)

Substitute our expressions for R(x) and C(x) and simplify:

P(x) = (-0.1x² + 525x) - (140x + 73500)

P(x) = -0.1x² + 525x - 140x - 73500

P(x) = -0.1x² + 385x - 73500

Now we can find the zero of the first derivative of this to find the "x" needed to  make the maximum profit:

P'(x) = -0.2x + 385

0 = -0.2x + 385

0.2x = 385

x = 1925

They have to sell \$1,925 units.  Now we do the same logic we did in the second part to find the selling price per item and then find the rebate needed from the \$420 purchase price:

p(x) = -0.1x + 525

p(1925) = -0.1(1925) + 525

p(1925) = -192.5 + 525

p(1925) = 332.5

That's the price after the rebate, so the amount of the rebate is:

420 - 332.5 = \$87.50