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## Review of Maximize Revenue Profit

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**Review of Maximize Revenue Profit**Ted mitchell**Finding the Price to Maximize Revenue**Ted Mitchell**Folklore and Beliefs**• Under normal circumstances: If I increase my selling price, then I will increase the amount of stuff I’ll sell? • True or False**FolkLore and Beliefs**• The optimal price in the new shoes game for maximizing profit is the price that maximizes revenue? • True or False**Folklore and Beliefs**• Under normal circumstances: In new shoes simulation game If I decrease my optimal selling price for making profit, then I will increase my sales revenue. • True or False**Folklore and Beliefs**• Cost based pricing is fair because it only bases along the increased cost of marking the product to the customer. • True or False**Folklore and Beliefs**• Demand Based pricing is NOT fair because it shifts the whole amount of an increase in cost plus a bigger profit margin to the customer. • True or False**Higher Price Sells Fewer Units**Quantity Sold Demand Equation Q = a - bP 100 ??? Price per Unit $50 $75 TJM**Get an estimate of a & b**Quantity Demand Equation Q = 1,500 -5P a =1,500 X 500 X X X X X X X X X Price per Unit $200 TJM**The Revenue Equation**• Revenue, R = P xQ • Q = a - bP Substitute Q = (a-bP) • R = P(a - bP) • R = aP - bP2**$5 Price Maximizes Revenue???R = 5,000P - 500P2**Revenue $12,500 $12,000 0 $4 $5 $6 Price TJM**Find the optimal Price that Maximizes Revenue**• Establish the Demand Function, Q = 5000 -500P • Establish the Revenue function • R = P x Q = P x (5000 -500P) = 5000P -500P2 • Find the first derivative wrt price • dR/dP = 5000 – 2(500)P = 5000 -1000P • Set the first derivative equal to zero and solve for P • 5000 – 1000P = 0 • P = 5000/1000 = $5**The maximum revenue is**• Revenue = P x Q = P x (5,000 – 500P) • Substitute optimal P = $5 • Max revenue = 5,000P – 500P2 • Max revenue = 5(5,000 – 500(52)) • Max Revenue = 25,000 – 12,500 = $12,500**The optimal price for max revenue**• The Demand For Your Product Changes with The Price You Charge As • Q = 5000 - 500P**$6 & $4 Examples**• The Demand For Your Product Changes with The Price You Charge AsQ = 5000 - 500P P = 5,000/2(500) = $5**Price For Maximizing Profit**by Ted Mitchell**Expand The Basic Profit Equation**Z = PQ - VQ - F substitute Q = ƒ(P) = a - bP Z = P(a-bP) - V(a-bP) - F Z = aP - bP2 - aV + bPV - F**Revenue looks likeR = aP - bP2With zero costs Revenue =**Profit Revenue = Profit 0 Price**Subtract Variable Costs from RevenueR - VQ - F = aP - bP2 -**aV + bPV- F Revenue = Profit 0 P* Price P***Example Exam Question**• The Demand is estimated by market research to beQ = 5,000 – 500PThe variable cost per unit is, V = $2The fixed cost for the period is, F = $7,000What is the selling price that will maximize the Profit? • First build the Profit Equation, ZThe Revenue is R = P(a-bP2) = P(5,000-500P) • The Profit is Z = R – VQ – F • Z = P(5,000-500P) – 2(5,000-500P) – 7,000 • Z = 5,000P – 500P2 -10,000+1000P – 7,000**Example Exam Question**• The Demand is estimated by market research to beQ = 5,000 – 500PThe variable cost per unit is, V = $2The fixed cost for the period is, F = $7,000What is the selling price that will maximize the Profit? • Second: Find the first derivative wrt P, • Z = 5,000P – 500P2 -10,000 + 1,000P – 7,000 • dZ/dP = 5,000 –2(500)P +1,000, set dZ/dP= 0 • 5,000 –2(500)P +1,000 = 0 , solve for P • –2(500)P = -5,000 -1,000 = P = 6,000/1,000 = $6**The optimal Price for maximizing profit in the example is**• Pz* = $6**Revenue**Profit $ $5 0 $6 Price**There is a General Solution for Finding Optimal Price for**Max Profit • Establish the Profit equation • Z = aP - bP2 - aV + bPV – F • Find the first derivative of the profit equation • dZ/dP = a – 2bP – bV • Set the first derivative equal to zero • dZ/dP = a – 2bP – bV = 0 • Solve for the optimal price • P = a/2b + bV/2b = a/2b + V/2b**The Price That Maximizes Profit**Consider Market Potential Consider Your Variable Costs Consider The Customer’s Sensitivity to Price Changes**The Price That Maximizes Profit**P = (Price that maximizes revenue) + (Half of the Variable Cost)