A manufacture has been selling 1750 television sets a week at $540 each. A market survey indicates that for each $14 rebate offered to a buyer, the number of sets sold will increase by 140 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.
p(x)=-.1x+715
b) How large rebate should the company offer to a buyer, in order to maximize its revenue?
=182.5
c) If the weekly cost function is 157500 + 180 x, how should it set the size of the rebate to maximize its profit?
the answers to A and B are right but i don't know how to do part c, could someone explain how to find part c?
You would solve c) in nearly the same method you used to solve b). Always always always, maximize provits by setting marginal cost (MC) = marginal revenue (MR). MR is the first derivitive of total revenue. Since you got b) right, you probably correctly calculated MR = 715 - .2x. (Then you maximized by setting this equal to zero and solving for x -- correct??)
For c) set MR = MC. MC is the first derivitive of the Total cost function, so MC is simply 180.
Take it from here.
how did you get part a
To find the size of the rebate that will maximize the company's profit, we need to take into account both the revenue and cost functions.
The revenue function can be calculated by multiplying the number of sets sold by the corresponding price, which is given by the demand function p(x). So the revenue function R(x) can be expressed as:
R(x) = x * p(x)
In this case, the weekly cost function C(x) is given as 157500 + 180x.
The profit function P(x) can be calculated by subtracting the cost function from the revenue function:
P(x) = R(x) - C(x)
To maximize profit, we need to find the value of x that maximizes P(x).
Now replace the functions with their respective expressions:
P(x) = [x * p(x)] - (157500 + 180x)
Simplify the expression by expanding:
P(x) = -0.1x^2 + (715 - 180)x - 157500
To find the value of x that maximizes the profit, we need to find the maximum point of this quadratic function.
The x-value of the maximum point of a quadratic function in the form ax^2 + bx + c can be found using the formula:
x = -b / (2a)
In this case, a = -0.1, b = (715 - 180), and c = -157500.
Plugging in the values, we have:
x = -[(715 - 180)] / (2 * -0.1)
= -[535] / (-0.2)
= -(-2675)
= 2675
Now that we have the value of x that maximizes the profit, we can find the size of the rebate by substituting x back into the demand function p(x):
p(x) = -0.1x + 715
p(2675) = -0.1 * 2675 + 715
= -267.5 + 715
= 447.5
Therefore, the company should set the size of the rebate to $447.5 in order to maximize its profit.
To find the size of the rebate that maximizes the company's profit, we need to understand how profit is calculated. Profit is equal to revenue minus cost: Profit = Revenue – Cost.
In part (b), we found that the revenue function is given by: Revenue (R) = x(p(x)), where x represents the number of television sets sold per week, and p(x) represents the price function.
In this case, the price function is p(x) = -0.1x + 715 (as you correctly calculated). So, the revenue function is obtained by multiplying the number of units sold (x) by the price (p(x)):
Revenue (R) = x(p(x)) = x(-0.1x + 715) = -0.1x^2 + 715x
Now, we also have the cost function, which is given as 157500 + 180x. To find the profit function, we subtract the cost function from the revenue function:
Profit (P) = Revenue (R) – Cost = (-0.1x^2 + 715x) – (157500 + 180x)
Simplifying this expression, we get:
P = -0.1x^2 + 715x - 157500 - 180x
Now, to maximize profit, we need to find the optimal value of x. This can be done by taking the derivative of the profit function and setting it equal to zero:
dP/dx = 0
Let's find the derivative of the profit function:
dP/dx = -0.2x + 715 - 180
Setting it equal to zero:
-0.2x + 715 - 180 = 0
Simplifying:
-0.2x + 535 = 0
-0.2x = -535
x = -535 / -0.2
x = 2675
So, the optimal value of x (the number of television sets sold per week) is 2675.
Now, we need to find the corresponding size of the rebate. To do this, we can plug this value of x back into the price function:
p(x) = -0.1x + 715
p(2675) = -0.1(2675) + 715
p(2675) = 182.5
Therefore, the company should offer a rebate of $182.5 to maximize its profit.