4. Give the demand function Q=100 - 0.2P, determine: a. The value of Q that maxi

Question

4. Give the demand function Q=100 - 0.2P, determine:

a. The value of Q that maximizes total revenue. Also, the price and the point elasticity at this output rate

b. If the firm is currently charging $60, should price be increased or decreased to increase total revenue? Explain

3. Market Equilibrium. Various beverages are sold by roving vendors at Busch Stadium, home of the St. Louis Cardinals. Demand and supply of the product are both highly sensitive to changes in the weather. During hot summer months, demand for ice-cold beverages grows rapidly. On the other hand, hot dry weather has an adverse effect on supply in that it taxes the stamina of the vendor carrying his or her goods up and down many flights of stairs. The only competition for this service is provided by the beverages that can be purchased at kiosks located throughout the stadium.
Demand and supply functions for ice-cold beverages per game are as follows:
Qd = 20,000 - 20,000P + 7,500Pk + 0.8Y + 500T (Demand)
Qs = 1,000 + 12,000P - 900Pl - 1,000Pc - 200T (Supply)

where P is the average price of ice-cold beverage ( $ per berverage), Pk is the average price of beverages sold at the kiosks ($ per beverage), Y is disposable income per household for baseball fans, T is the average daily high temperature (Degrees), Pl is the average price of unskilled labor ($ per hour), and Pc is the average cost of capital (in percent)

Q)- Calculate the market equilibrium price-output combination.

2. The Market research department at a nursery determines that demand for garden shrubs is given by:
Q= 1,000 - 5Ps +0.051 - 50Pt
where Ps is the price of shrubs, I is income, and Pr is the price of trees.
If Ps= $5, I=$20,000, and Pt=15,

c. Compute the cross elasticity for shrubs and trees. Are these two goods substitutes or complements?

Explanation / Answer

R(x)=xD(x) R(x)=100x-.2x^2 (4A)w/calculus dR(x)/dx=-.4x+100 ; the critical point is 250; this is where the rate of change = 0, and for parabolas when the rate of increase becomes negative. wo/ calculus -b/2a to find the top of the parabola which is -100/-.4=250 Therefore the maximum possible revenue is R(250)=100(250)-.2(250)^2 =$12,500 To calculate the optimal price, plug the optimal revenue point (250) into the demand function D(250)=100-.2(250)=$50 (4B)Short answer is : $50 is the optimal price, any other price will provide less revenue. Long answer :The optimal revenue will be obtained at $50. We need to calculate where on the parabola $60 is so we can compare the difference in revenue. D(x)=60=100-0.2x. basic algebra shows D(200)=$60. So only 200 units will be sold at $60 compared to the 250 at $50. R(200)=100x-.2x^2=$12000 in revenue. This is $500 less than revenue than if the price was $60. (3)You need to find where Qd=Qs. This takes some basic algebra so you can solve for the price and solve for the output. (2c) You need to find the derivative of Ps in respect to Pt. If the ratio is positive they are complimentary, negative substitutes

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