1. The demand function is the function that gives quantity demanded as a function of price. Assuming demand is linear, find the demand function for these electric chainsaws.
Q=m(200) + 200
Y=mx+b à 200=m(14,000)+b
M= -150, so à p=-150x+b
2. Given this demand function, what is the largest number of chainsaws that can be sold?
Maximum revenue from demand function = Rp = 14,000-1/150Q;
0=14,000-2/150Q, Max revenue therefore = 150*7,000 = $1,050,000
3. Once you know the demand function and a fixed number of chainsaws is produced, you will set the price as large as possible. What is the revenue function? That is, give the amount of money you make as a function of the quantity produced.
Price would be for 14,000 units, $150.
4. Find the profit function.
P(x) = R(x) – C(x)
Revenue = -150x2+7000x
Px= 150x2+7000x – (95x+14,000)
150x2+7000x – 95x-14,000
5. If the current production is 5,000 chainsaws per year, how much extra profit would the company get if they produced one more chainsaw?
P(5,000) = 150(5,000)2+6905(5,000)-14,000
6. Based on the increase in profit for one additional chainsaw, at a level of 5,000, would you recommend producing more? Why?
MR= Change in Total Revenue/Change in quantity sold MR= 5000@$200, 5001@$199.5
MR = $2
7. If the current production were 20,000 chainsaws per year, how much extra profit would the company get if they produced one more chainsaw?
Profit maximization point is when MR=MC
20,000*$200 – 20,000*$95 OR 20,000(105) – 400,000 = 1,700,000
19,896 if 20,001 units are produced.
8. Based on the increase in profit for one additional chainsaw, at a level of 20,000, would you recommend producing more? Why?
Marginal cost is generated only when an additional unit is produced (or there is an increase in the unit of production by ONE unit). This means that with the production of chainsaw will result in increase in demand due to lower price per additional unit produced, thus ensuring profit maximization during short run. The profit maximization will only until MR exceeds MC. The profit maximization will end until MC equates MR. So with 20,001 units there will be economies of scale, reduced price and more demand.
9. Explain how analyzing the additional profit of one more chainsaw relates to calculus.
Calculus is regularly used in economics, especially derivatives. For instance, in this case, economists will use calculus to determine the correct time to sell chainsaw, using the actioner marginal value concept – ascertaining the price of chainsaw affects the people’s perception to buy it or measuring specific changes overtime in respect to one or two variables. This is what we call derivatives.
10. Using the previous problems where we analyzed the additional profit of producing one more chainsaw, how would you decide the number of chainsaws to product to maximize profit?
IF it is in a perfect competition they will be price taker. So the revenue = market price at given chainsaw production. The flat or linear demand function shows either increase or decrease from relatively low quantity or higher chainsaw. If they are in monopoly, it will equate market demand curve. For profit maximization (where MR=MC), the number of chainsaw should be at $2 for each unit.
11. The relative rate of change of a function at a point is the rate of change at that point divided by the value of the function at that point. Find the relative rate of change of profit.
Average rate of change = f(x) = x2
f(b)-f(a)/b-a, if the interval is 1 and 4
12. The elasticity of demand is the negative relative rate of change of change of demand, divided by the relative rate of change of price. Give an intuitive interpretation of what elasticity of demand means.
Elastic demand curve depicts the steeper manner where the change in price (such as from $200 to $220) resulted in greater change in demand in response (from 14,000 units to 11,000 units). It is an elastic demand where the response to change in demand is higher than change in price (that rouse by $20).
13. Find the elasticity of demand as a function of price.
ǫ = p/ q dq/dp
dp/dq<0, implying ǫ to be –ve.
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