Maximum profit?

[NikkiGurl.]

A company manufactures x calculators weekly that can be sold for 75-0.01x dollars each at a cost of
1850+28x-x^2+.001x^3 for manufacturing all x calculators. What is the number of calculators the company should manufacture in order to maximize its weekly profit?

Can someone please help me start this? I'm sure I can figure it out if i just knew where to start..

 

[fasteddie65]

A company manufactures x calculators weekly that can be sold for 75-0.01x dollars each at a cost of
1850+28x-x^2+.001x^3 for manufacturing all x calculators. What is the number of calculators the company should manufacture in order to maximize its weekly profit?

Can someone please help me start this? I'm sure I can figure it out if i just knew where to start..

The profit equals the income minus the expense.

The income is (75 - 0.01x)x = 75x - 0.01x^2.

The expense is 1850 + 28x - x^2 + .001x^3.

The profit is 75x - 0.01x^2 - (1850 + 28x - x^2 + 0.001x^3) = -1850 - 47x - 1.01x^2 - 0.001x^3.

To maximize the profit, set the derivative -47 - 2.02x - 0.003x^2 = 0.

 

[BigGlenntheHeavy]

x = number of units produced

p = price per unit = 75-.01x

R = total revenue from selling x units, R = xp

C = total cost of producing x units, C = 1850+28x-x^2+.001x^3

P = total profit from selling x units, P = R-C

Hence P = x(75-.01x)-(1850+28x-x^2+.001x^3)

Find dP/dx, set it to zero and solve for x and that will give you the max number of units you should sell to max your profit.

 

[BigGlenntheHeavy]

An afterthought. Supposes you own a cab company with 100 cabs and you figured out with equations that your profit was maxamize when 80 of your 100 cabs were on the street. Now the Shriners come into town and all 100 of your cabs are busy.
Hence all your friencd say, hey he must be really raking in the bucks, not one of his cabs is idle, when in fact your profit is less that when you had only 80 cabs in service.

Food for thought.