Question

A manufacturer sells lamps at six dollars each and sells 3000 each month. For each one dollar that the price is increased, 1000 fewer lamps are sold each month. It costs 4 dollars to make one lamp. What price should lamps be sold at to maximize profit

Answer 1

Answer:

Price it will be sold to make maximum profit = $6.5 each

Explanation:

The challenge is to maximize profit.

We can assign $x$ to be the number of dollars obtained per unit price per flashlight.

Recall:

Profit = Revenue - Cost

Revenue = price charged x number sold =$(6 + x) (3000 - 1000x)$

Cost = cost to produce a unit x number of units produced $= (4)(3000-1000x)$

$(6 + x)(3000 - 1000x) - (4)(3000 - 1000x)$

Profit = $1000x^{2} + 1000x +6000$

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

$\frac{dP}{dx}= -2000x +1000=0$

∴ $x=0.5$

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

Answer 2

Answer:

Explanation:

Given:

Selling price of 1 lamp = $6

Cost price of 1 lamp = $4

Units sold per month = 3000

Let $T be the selling price set by the lamp seller.

Number of sold lamps per month = 3000 − (T − 6) × 1000

= 9000 − 1000 × T.

Monthly profit = (9000 − 1000p) × (T − 4)

= −1000T^2 + 13000T − 36000.

Obtaining the derivative,

dS/dT = −2000T + 13000

and setting it to zero

−2000T + 13000 = 0

T = -13000/-2000

optimal selling point, T = $6.5.