A manufacture has been selling 1150 television sets a week at $510 each. A market survey indicates that for each $13 rebate offered to a buyer, the number of sets sold will increase by 130 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?
If the weekly cost function is 97750 + 170 x, how should it set the size of the rebate to maximize its profit?
a) To find the function representing the demand, p(x), we need to determine the relationship between the number of sets sold (x) and the corresponding price (p(x)).
We are given that the manufacturer has been selling 1150 television sets a week at $510 each. This means that when x = 1150, p(x) = $510.
We are also given that for each $13 rebate offered to a buyer, the number of sets sold will increase by 130 per week. This implies a linear relationship between the rebate and the increase in sets sold.
Let's denote the rebate as r and the increase in sets sold as i(r). Based on the given information, we can construct the following linear relationship:
i(r) = 130(r/13)
This equation states that for every $13 increase in rebate (r), the number of sets sold (i) increases by 130.
To find p(x), we can use the equation:
p(x) = 510 + i(r)
Since i(r) = 130(r/13), we can substitute this into the equation:
p(x) = 510 + 130(r/13)
Simplifying:
p(x) = 510 + 10r
Therefore, the function representing the demand is p(x) = 510 + 10r, where x is the number of television sets sold per week and p(x) is the corresponding price.
b) To maximize revenue, we need to determine the rebate that will result in the highest total revenue.
Revenue (R) is calculated by multiplying the price (p(x)) by the quantity of sets sold (x):
R = x * p(x)
Substituting p(x) = 510 + 10r into the equation:
R = x * (510 + 10r)
To find the rebate that maximizes revenue, we can take the derivative of R with respect to r and set it equal to zero:
dR/dr = x * 10 = 0
From this equation, we can see that x must equal zero or be undefined for the derivative to be zero. However, this doesn't make sense in the context of the problem. Therefore, the derivative will not help us find the maximum.
Instead, we will use a different approach to find the optimal rebate.
To find the rebate that maximizes revenue, we can plot a graph of R in terms of r and identify the highest point.
Using the given cost function of 97750 + 170x, we can calculate the profit function:
Profit (P) = Revenue (R) - Cost (C)
P = R - C
P = x * p(x) - (97750 + 170x)
Now we can substitute p(x) = 510 + 10r into the equation:
P = x * (510 + 10r) - (97750 + 170x)
Simplifying:
P = -160x + 10rx - 97750
To find the rebate that maximizes the profit, we need to differentiate P with respect to r and set it equal to zero:
dP/dr = 10x - 160 = 0
Solving for x:
10x = 160
x = 16
Now that we have the optimal value of x, we can substitute it back into the profit function to find the rebate:
P = -160x + 10rx - 97750
P = -160(16) + 10r(16) - 97750
P = -2560 + 160r - 97750
P = 160r - 100310
Therefore, to maximize profit, the company should set the size of the rebate at $160.