X articles are produced at a total cost of $(100-30x+x^2) and each is sold for $3x/2. How many should be made to produce the largest profit and how much is the profit
P = 3 x^2 / 2 - 100 + 30 x - x^2
P = .5 x^2 + 30 x - 100
find vertex or use calculus
dP/dx = x + 30
= 0 at max so x is - 30 for max or min. Typo.
To find the number of articles that should be made to produce the largest profit, we need to maximize the profit function. The profit function can be calculated by subtracting the cost from the revenue.
Revenue = Selling price per article * Number of articles
R(x) = (3x/2) * x = 3x^2/2
Cost = Total cost
C(x) = (100 - 30x + x^2)
Profit = Revenue - Cost
P(x) = R(x) - C(x) = 3x^2/2 - (100 - 30x + x^2)
To find the number of articles that should be made to maximize profit, we need to find the critical points of the profit function. Critical points occur where the derivative of the profit function is zero or does not exist.
Let's find the derivative of P(x) with respect to x:
P'(x) = d/dx (3x^2/2 - (100 - 30x + x^2))
= 3x - (30 - 2x)
= 5x - 30
To find the critical points, we set the derivative equal to zero and solve for x:
5x - 30 = 0
5x = 30
x = 6
Now, we need to verify if this is a maximum or minimum point. We take the second derivative of the profit function and evaluate it at x = 6.
P''(x) = d^2/dx^2 (3x^2/2 - (100 - 30x + x^2))
= 3 - 0
= 3
Since the second derivative is positive, the critical point x = 6 is a minimum point. Therefore, it is the point that maximizes the profit.
To find the actual profit, we substitute x = 6 into the profit function:
P(6) = 3(6)^2/2 - (100 - 30(6) + (6)^2)
= 54 - (100 - 180 + 36)
= 54 - (-116)
= 170
Hence, the maximum profit can be obtained by producing 6 articles, and the profit at this quantity is $170.