Answer :
a. Substituting the given price equation p = 280x - 0.4x^2
b. The total profit= -1x^2 + 280x - 5000
c. 140 ovens must be produced and sold in order to maximize profit.
d. The coefficient of the x^2 term is -1, which is negative, the parabola opens downward and the profit function has a maximum.
e. The maximum profit is $14,600.
f. The price per oven that maximizes profit is $224.
A. The total revenue, R(x), can be calculated by multiplying the number of ovens, x, by the price per oven, p:
R(x) = x * p
Substituting the given price equation p = 280 - 0.4x:
R(x) = x * (280 - 0.4x)
= 280x - 0.4x^2
B. The total profit, P(x), is the difference between total revenue and total cost:
P(x) = R(x) - C(x)
= (280x - 0.4x^2) - (5000 + 0.6x^2)
= 280x - 0.4x^2 - 5000 - 0.6x^2
= -1x^2 + 280x - 5000
C. To find the number of ovens that maximize profit, we need to find the vertex of the quadratic function P(x). The x-coordinate of the vertex can be found using the formula -b/2a, where a = -1 and b = 280:
x = -280 / (2*(-1))
x = -280 / (-2)
x = 140
Therefore, 140 ovens must be produced and sold in order to maximize profit.
D. To prove that the profit is at a maximum, we can examine the concavity of the profit function P(x). Since the coefficient of the x^2 term is -1, which is negative, the parabola opens downward and the profit function has a maximum.
E. To find the maximum profit, we substitute the value of x = 140 into the profit function P(x):
P(140) = -1(140)^2 + 280(140) - 5000
Simplifying the expression, we get:
P(140) = -19600 + 39200 - 5000
= 14600
Therefore, the maximum profit is $14,600.
F. The price per oven that maximizes profit can be found by substituting the value of x = 140 into the price equation p = 280 - 0.4x:
p = 280 - 0.4(140)
= 280 - 56
= 224
Therefore, the price per oven that maximizes profit is $224.
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