For a monopolist’s product, the demand function is 𝑝 = 72 − 0.04𝑞 and the cost
function is 𝑐 = 500 + 30𝑞. At what level of output will the profit be maximized? At
what price does this occur and what is the profit?
ANSWER: 525, RM51, RM10,525
I WANT THE STEPS
ITS OPTIMIZATION
Well, well, well, we have ourselves an optimization problem, don't we? Let me put on my optimization glasses and solve this for you step by step:
Step 1: We need to find the total revenue function. Total revenue (TR) is calculated by multiplying price (p) by quantity (q). So, TR = p * q. Given that the demand function is p = 72 - 0.04q, we can substitute this equation into the total revenue equation: TR = (72 - 0.04q) * q.
Step 2: Now, we need to find the total cost function. Total cost (TC) is given as c = 500 + 30q.
Step 3: The profit (π) is calculated by subtracting total cost from total revenue: π = TR - TC. Let's substitute our equations from Step 1 and Step 2 into this equation: π = (72 - 0.04q) * q - (500 + 30q).
Step 4: Simplify the profit equation by multiplying out the terms: π = 72q - 0.04q^2 - 500 - 30q.
Step 5: Rearrange the equation in descending order of the degree of q: π = -0.04q^2 + 42q - 500.
Step 6: To find the level of output where profit is maximized, we need to find the value of q that maximizes the profit. This is done by taking the derivative of the profit equation and setting it equal to zero. So, let's find the derivative of the profit equation: π' = -0.08q + 42.
Step 7: Set the derivative equal to zero and solve for q: -0.08q + 42 = 0. Solving for q gives us q = 525.
Step 8: Now that we have the level of output, we can substitute this value back into the demand function to find the price. Substituting q = 525 into p = 72 - 0.04q gives us p = 72 - 0.04 * 525 = 72 - 21 = RM51.
Step 9: To find the profit at the maximum level of output, substitute q = 525 into the profit equation π = -0.04q^2 + 42q - 500: π = -0.04 * (525)^2 + 42 * 525 - 500 = -0.04 * 275625 + 22050 - 500 = -11025 + 22050 - 500 = RM10,525.
And there you have it! The level of output where profit is maximized is 525 units, the price at this level is RM51, and the profit is RM10,525. Optimized and calculated with a touch of humor!
To find the level of output at which profit is maximized, we need to determine the quantity that maximizes the difference between total revenue and total cost. Here's a step-by-step approach:
Step 1: Calculate total revenue (TR):
Since the price (p) is given as a function of quantity (q), we can substitute the demand function p = 72 - 0.04q into the equation for total revenue.
TR = p * q
TR = (72 - 0.04q) * q
TR = 72q - 0.04q^2
Step 2: Calculate total cost (TC):
The cost function is given as c = 500 + 30q, so we can substitute this into the equation for total cost.
TC = c * q
TC = (500 + 30q) * q
TC = 500q + 30q^2
Step 3: Calculate profit (π):
Profit is equal to total revenue minus total cost.
π = TR - TC
π = (72q - 0.04q^2) - (500q + 30q^2)
π = 72q - 0.04q^2 - 500q - 30q^2
π = -30.04q^2 + 72q - 500q
Step 4: Find the derivative of the profit function (dπ/dq):
To find the quantity at which profit is maximized, we differentiate the profit function with respect to q and set it equal to zero.
dπ/dq = -60.08q + 72 - 500 = 0
-60.08q = 428
q = 428 / -60.08
q ≈ -7.12
Step 5: Determine the level of output at which profit is maximized:
The level of output cannot be negative in this case since it is a physical quantity. Therefore, the monopolist should produce a positive quantity of units.
So, we can conclude that at a level of output of approximately 7.12 units, the profit is maximized.
Step 6: Calculate the price at which this occurs:
To find the price at which the profit is maximized, substitute the quantity value (7.12) into the demand function p = 72 - 0.04q.
p = 72 - 0.04(7.12)
p ≈ 72 - 0.2848
p ≈ 71.72
Step 7: Calculate the profit at the maximum level of output:
To find the maximum profit, substitute the quantity value (7.12) into the profit function π = -30.04q^2 + 72q - 500q.
π = -30.04(7.12)^2 + 72(7.12) - 500(7.12)
π ≈ -30.04(50.6944) + 511.68 - 3560
π ≈ -1522.78 + 511.68 - 3560
π ≈ -2552.10
The profit at the level of output where it is maximized is approximately -2552.10 (negative since the cost is higher than the revenue).
Therefore, at a level of output of approximately 7.12 units, the price at which this occurs is approximately $71.72, and the profit is approximately -$2552.10.
To find the level of output at which profit is maximized, we need to maximize the profit function. The profit function is given by:
Profit = Total Revenue - Total Cost
To find the total revenue, we multiply the price, denoted as p, by the quantity sold, denoted as q. So the total revenue function is:
Total Revenue = p * q
On substituting the demand function p = 72 - 0.04q, we get:
Total Revenue = (72 - 0.04q) * q
Expanding the equation, we have:
Total Revenue = 72q - 0.04q^2
The total cost function is given as c = 500 + 30q.
Now, we can write the profit function as:
Profit = Total Revenue - Total Cost
= (72q - 0.04q^2) - (500 + 30q)
= -0.04q^2 + 42q - 500
To find the level of output where profit is maximized, we need to find the derivative of the profit function with respect to q and set it equal to 0. So:
Profit'(q) = 0
-0.08q + 42 = 0
Solving the equation, we get:
q = 525
Thus, the level of output at which profit is maximized is 525 units.
To find the price at which this occurs, we substitute the value of q = 525 into the demand function p = 72 - 0.04q:
p = 72 - 0.04 * 525
= 72 - 21
= 51
Therefore, the price at which the profit is maximized is RM51.
To find the maximum profit, we substitute the value of q = 525 into the profit function:
Profit = -0.04 * (525)^2 + 42 * 525 - 500
= -1099 + 22050 - 500
= RM10,525
Hence, the maximum profit is RM10,525.
I don't see anything about price. Is p(q) the price? If so, then since revenue = price * quantity,
profit = pq-c
so expand that and find where the derivative is zero.