Question

Price-Demand Equation x = 2750-25p

Fixed Cost 8000

Variable Cost 50

Construct the cost function C(x) and describe the monthly cost of this business.

Construct the profit function R(x) and describe the monthly revenue. Determine the domain of R(x) which represents the range of units it could produce.

Construct the profit function P(x), determine break-even points, what production levels are profitable, what levels incur a loss. graph this.

at a monthly level of production within the domain of the revenue function (you can select within the profitable amount), determine the total cost, revenue, and profit at this level. Then determine the marginal cost, marginal revenue and marginal profit plus interpret these.

According to the price – demand equation, at what unit price ($p)) are you selling your product if demand is at your chosen production level? Write a function for the elasticity of demand E(p). Is E(p) elastic or inelastic, why? how increasing or decreasing the price would affect revenue. What unit price would result in unit elasticity?

What is the optimal production level that will maximize profits and find the max profit, show equation and graph.

Determine the revenue and cost at this optimal production level. Use the price – demand equation to determine what price should you sell each unit so that you can maximize profit.

Answer 1

Here the cost function C(x) = 8000 + 50x

Profit function R(x) = p* x = x * (2750-x)/25

Here the domian of monthluy productionis from x = 0 to x = 2750 as on both values, production is nil.

P(x) = R(x) - C(x) = (2750 - x)x/25 - (8000 + 50x)

Here at breakeven points. profit shall be maximum or say,

dP(x)/dx = 0

dP/dx = 1/25 * (2750 - x -x) - 50 = 0

2750 - 2x - 1250 = 0

2x = 1500

x = 750 units

Here maximum profit will be at x = 750 units.

Here breakeven point is where P(x) = 0

(2750 - x)x/25 - (8000 + 50x) = 0

(2750 - x)x/25 = (8000 + 50x)

2750x - x²= 200000 + 1250 x

x²- 1500x + 200000 = 0

x = 147.92 and x = 1352.08

so here for values of x < 147.92 and x > 1352.08, profit is negative and for 147.92 < x < 1352.08, profit is positive.

Here monthly level of production x = 1000

C(x) = 8000 + 50 * 1000 = 58000

R(x) = 70000

P(x) = 12000

Marginal Cost = MC (x = 1000)= dC/dx = 50

It is the cost incrersad when we increase thwe prioduction by 1 unit.

Marginal Revenue MR = dR/dx =  1/25 * (2750 - 2x)

so MR(x = 1000) = 1/25 * (2750 - 2 * 1000) = 30

so here marginal revenue means that if we increase the quantitiy by 1 unit, then it will increase the revenue by the value of 30 units.

Here marginal profit = MP = DP/dx = 1/25 * (2750 - 2x) - 50

so MP(x = 1000) =-20

so at production level x = 1000, if we increase the prodcution level by 1 unit, it will decrease the profit by -20.

Here price at x= 1000

P(x = 1000) = (2750 - 1000)/25 = 70

E[p] = (dx/Q)/ (dx/P) = (dx/dP) * (P/x)

E[p] = -25(2750 - x)/25x = -(2750 - x)/x

E[p] = (x - 2750)/x

so at x = 1000

E[p] = (1000 - 2750)/1000 = -1.75

so here the demand is very elastic

so there be unit elasticity when

(2750 - x) = x

2750 = 2x

x = 2750/2 = 1375

so x = 1375, demand eslasticity would be unit e\lasticity,.

Here optimum prodcution level is x = 750

max profit = (2750 - 750) * 750/25 - (8000 + 50 * 750) = 14500

Revenue at x = 750 = (2750 - 750) * 750/25 = 60000

Cost (x = 750) = 8000 + 50 * 750 = 45500

Here p =- (2750 - 750)/25 = 2000/25 = 80