Respuesta :
Answer:
Price it will be sold to make maximum profit = $6.5 each
Explanation:
The challenge is to maximize profit.
We can assign [tex]x[/tex] to be the number of dollars obtained per unit price per flashlight.
Recall:
Profit = Revenue - Cost
Revenue = price charged x number sold =[tex](6 + x) (3000 - 1000x)[/tex]
Cost = cost to produce a unit x number of units produced [tex]= (4)(3000-1000x)[/tex]
[tex](6 + x)(3000 - 1000x) - (4)(3000 - 1000x)[/tex]
Profit = [tex]1000x^{2} + 1000x +6000[/tex]
At the point of maximum profit, the change in profit with respect to price will be = 0. In mathematical terms, the derivative of the profit will be = zero
[tex]\frac{dP}{dx}= -2000x +1000=0[/tex]
∴ [tex]x=0.5[/tex]
from this, we can see that at maximum profit, x = 0.5
∴ Price it will be sold to make maximum profit = $6.00 + 0.5 = $6.5 each
The price that the lamps should be sold to maximize profit is $6.50.
The profit will be calculated as:
= Revenue - Cost
= (6 + x)(3000 - 1000x) - 4(3000 - 1000x)
= 1000x² + 1000x + 6000
Then, we'll differentiate profit which will be:
- 2000x + 1000 = 0
- 2000x = 0 - 1000
- 2000x = -1000
x = -1000 / -2000
x = 0.5
Therefore, the maximum profit will be:
= $6.00 + 0.50 = $6.50
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