Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

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?

1 Answer

Relevance
  • 1 month ago
    Favorite Answer

    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

Still have questions? Get your answers by asking now.