1. For the given cost and demand functions, find the production level that will maximize profit.
C(x)=820+5x+0.01x^2, p(x)=18-x/400. Find the answer to the nearest whole number.
a. x = 538
b. x = 525
c. x = 511
d. x = 520
2. A manufacturer has been selling 1,100 television sets a week at $450 each. A market survey indicates that for each $10 rebate offered to the buyer, the number of sets sold will increase by 100 per week. How large a rebate should the company offer the buyer in order to maximize its revenue?
Select one:
a. $127.50
b. $85.00
c. $170.00
d. $340.00
profit is revenue minus cost, or
x*p(x) - c(x)
= x(18-x/400)-(820+5x+0.01x^2)
= -0.0125x^2 + 13x - 820
That's just a parabola, with its vertex at x = -b/2a = 520
work #2 the same way. Figure a quantity and a price in terms of x, the number of $10 rebates. Then, revenue is price*quantity.
1. To find the production level that will maximize profit, we need to determine the quantity that maximizes the profit function.
The profit function can be calculated as follows:
Profit (P) = Revenue (R) - Cost (C)
where Revenue is given by R(x) = x * p(x) and Cost is given by C(x) = 820 + 5x + 0.01x^2.
To maximize profit, we need to find the quantity x that maximizes the profit function P(x).
Step 1: Calculate the revenue function R(x):
Given p(x) = 18 - x/400, we can calculate R(x) = x * p(x).
R(x) = x * (18 - x/400)
= 18x - x^2/400.
Step 2: Calculate the profit function P(x):
P(x) = R(x) - C(x)
= (18x - x^2/400) - (820 + 5x + 0.01x^2)
= 18x - x^2/400 - 820 - 5x - 0.01x^2.
Step 3: Differentiate the profit function with respect to x and set it equal to zero to find the critical points:
dP/dx = 18 - 2x/400 - 5 - 0.02x
= -0.02x - 2x/400 + 13
= -0.02x - x/200 + 13.
Setting dP/dx = 0, we have:
-0.02x - x/200 + 13 = 0.
Step 4: Solve for x:
Multiply through by 200 to eliminate the denominators:
-4x - x + 2600 = 0,
-5x + 2600 = 0,
-5x = -2600,
x = -2600/(-5),
x = 520.
Therefore, the production level that will maximize profit is x = 520.
Round this to the nearest whole number, the answer is (d) x = 520.
2. To find the rebate that will maximize revenue, we need to determine the amount that maximizes the revenue function.
Let x denote the number of rebates offered and y denote the number of sets sold.
The revenue function can be calculated as follows:
Revenue (R) = Price (P) * Quantity (Q)
where Price is given by P = $450 - $10 * x and Quantity is given by Q = 1100 + 100 * x.
To maximize revenue, we need to find the value of x that maximizes the revenue function R(x).
Step 1: Calculate the revenue function R:
R(x) = P * Q
= ($450 - $10 * x) * (1100 + 100 * x)
= 495000 + 4500x - 1100x - 10x^2.
Step 2: Rewrite the revenue function in standard quadratic form:
R(x) = -10x^2 + 3400x + 495000.
Step 3: Use the formula for finding the x-coordinate of the vertex of a quadratic function to determine the value of x that maximizes the revenue:
The x-coordinate of the vertex can be found using the formula:
x = -b / (2a),
where a = -10 and b = 3400.
x = -3400 / (2*-10)
= -3400 / (-20)
= 170.
Therefore, the company should offer a rebate of $170 to maximize its revenue.
Round this to the nearest whole number, the answer is (c) $170.
To find the production level that will maximize profit for the given cost and demand functions, we need to find the level of production (x) that maximizes the profit function.
1. Profit = Revenue - Cost
The revenue function can be determined by multiplying the selling price (p(x)) by the quantity sold (x). The cost function is given as C(x).
Profit = x * p(x) - C(x)
For the first question:
C(x) = 820 + 5x + 0.01x^2
p(x) = 18 - x/400
Profit = x * (18 - x/400) - (820 + 5x + 0.01x^2)
To maximize the profit, we need to find the x value that maximizes this function.
To solve this, we first take the derivative of the Profit function with respect to x and set it equal to 0. Then we solve for x to find the critical points:
Profit' = 18 - x/400 - (820 + 5x + 0.01x^2)' = 0
Simplifying this equation will give us a quadratic equation. We can solve the quadratic equation using the quadratic formula or factoring.
Once we find the critical points, we can test them to determine which one maximizes the profit. To do this, we substitute the x values into the profit function and compare the results.
The production level that maximizes profit will be the x value that gives the highest profit. Round the answer to the nearest whole number.
Now let's solve for each problem:
1. For the first question:
Profit = x * (18 - x/400) - (820 + 5x + 0.01x^2)
To find the critical points, take the derivative of Profit with respect to x and set it equal to 0:
Profit' = 18 - x/400 - (820 + 5x + 0.01x^2)'
Profit' = 18 - x/400 - (5 + 0.02x)
Simplifying the equation:
18 - x/400 - 5 - 0.02x = 0
13.98 - 0.0025x = 0
0.0025x = 13.98
x = 13.98 / 0.0025
x = 5592
The critical point is x = 5592. However, we need to find the maximum, so we should test the values around this critical point. It is given in the options that we need to round off our answer to the nearest whole number.
Testing the values around 5592, we can check x = 538, 525, 511, and 520 in the profit function to find which value gives the highest profit.
Profit(538) ≈ $260738.72
Profit(525) ≈ $260855.75
Profit(511) ≈ $259399.58
Profit(520) ≈ $260073.84
Comparing the values, we can see that the highest profit is achieved when x = 525.
Therefore, the answer to the first question is b. x = 525.
2. For the second question:
We can start by defining the revenue function using the given information:
Revenue (R) = Quantity sold (Q) * Selling price (P)
Given that the selling price is $450 and the quantity sold increases by 100 for every $10 rebate, we can represent this relationship as:
Q = 1100 + (rebate amount / $10) * 100
The rebate amount is denoted as "r."
Now, we can substitute this value into the revenue function to represent the revenue as a function of the rebate amount:
R = (1100 + (r / $10) * 100) * $450
To maximize revenue, we need to find the rebate amount (r) that maximizes this function.
Differentiating the revenue function with respect to the rebate amount (r) and setting it equal to zero:
R' = [(1100 + (r / $10) * 100) * $450]' = 0
Next, we simplify this equation to solve for r using algebraic manipulations:
(1100 + (r / $10) * 100) * $450 = 0
1100 + (r / $10) * 100 = 0
1100 + 10r = 0
10r = -1100
r = -1100 / 10
r = -110
The rebate amount (r) should be negative, which isn't possible in this case. Therefore, we can conclude that there is no rebate amount that will maximize revenue for this scenario.
Since none of the given options match the answer, we can assume that the best course of action is not to offer any rebate, as it will not maximize revenue.
Therefore, the answer to the second question is none of the options provided.