Patrick D. answered • 04/26/17

Patrick the Math Doctor

Nora B.

asked • 04/26/17An air freight company has determined that the cost, in dollars, of delivering parcels per flight is C(x)=2025+7x The price per parcel, in dollars, the company charges to send x parcels is P(x)=22-0.01X Determine: The revenue function R(x)= The profit function P(x)= The company's maximum profit. The Prices per parcel that yields the maximum profit. The minimum number of parcels the air freight company musty ship to break even. (Revenue=Cost)

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Patrick D. answered • 04/26/17

Patrick the Math Doctor

(a) Revenue = R(x) = x * p(x) = 22x - 0.01X^2

(b) Profit = P$(x) = R(x) - C(x) = 22x - 0.01X^2 - (2025 + 7x)

= 22X - 0.01X^2 - 2025 - 7x

= -0.01X^2 + 15X - 2025

(c)

finding where derivative is zero : 0 = dP$/dx = -0.02X + 15

max profit at X = 750 parcels reaps $3600

Claretta W. answered • 04/26/17

Multi-disciplinary Instructor (Online and In-person)

Using 4025 + 5x instead of 2025+7x

The revenue, or how much money you collect is the number of parcels times the price per parcel.

So, R(x) = x*p(x) = x(20-.01x) = 20x - .01x^2

The profit is income - cost.....so, P(x) = R(x) - C(x) = 20x - .01x^2 - (4025 + 5x). Simplifying you'd get

P(x) = -.01x^2 +15x - 4025.

If you know calculus, take the derivative, set it = 0, and you'll find the x that leads to the maximum. Since this is labeled college algebra, I will do this with the properties of a parabola. The maximum (or minimum) value of a parabola is always on the axis of symmetry. You can tell if it's a maximum (or minimum) by looking at the coefficient of the x^2 term. Since our value is negative (-.01), we have a "frowning" parabola and so we do have a maximum. (negative frown but positive smile......corny but efficient)

Axis of symmetry = -b/2a or x= -15/-.02 =750. So 750 parcels leads to the maximum profit.

so maximum profit is P(750)= -.01(750)^2+15(750)- 4025 or 1600 dollars.

price per parcels is p(750) = 20-.01(750) = 12.50 dollars per parcel.

Break even point is when cost = revenue or 4025 +5x = 20x - .01x^2 .

402500 + 500x =2000x -x^2

x^2 - 1500x +402500 = 0

(x-350)(x-1150) = 0

x=350 x=1150

So, R(x) = x*p(x) = x(20-.01x) = 20x - .01x^2

The profit is income - cost.....so, P(x) = R(x) - C(x) = 20x - .01x^2 - (4025 + 5x). Simplifying you'd get

P(x) = -.01x^2 +15x - 4025.

If you know calculus, take the derivative, set it = 0, and you'll find the x that leads to the maximum. Since this is labeled college algebra, I will do this with the properties of a parabola. The maximum (or minimum) value of a parabola is always on the axis of symmetry. You can tell if it's a maximum (or minimum) by looking at the coefficient of the x^2 term. Since our value is negative (-.01), we have a "frowning" parabola and so we do have a maximum. (negative frown but positive smile......corny but efficient)

Axis of symmetry = -b/2a or x= -15/-.02 =750. So 750 parcels leads to the maximum profit.

so maximum profit is P(750)= -.01(750)^2+15(750)- 4025 or 1600 dollars.

price per parcels is p(750) = 20-.01(750) = 12.50 dollars per parcel.

Break even point is when cost = revenue or 4025 +5x = 20x - .01x^2 .

402500 + 500x =2000x -x^2

x^2 - 1500x +402500 = 0

(x-350)(x-1150) = 0

x=350 x=1150

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