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Nt1310 Unit 4 Case Study

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Recently I started up a business that manufacture phone cases to sell. The production cost for the unit $20. Public- service and rent for the total fixed cost is $2,750 per calendar month. Using the fixed cost to construct a linear function C(x)=8,000+20x.The given price- demand equation is x=f(p)=2750-25p, where x represents the quantity of units of phone cases needed each month and let p represent the price per unit in dollars. To solve for the price x=2750-25p x-2750=25p
-(x-2750)/25÷-25p/25
-0.04x+110= p
Marginal cost function will be C’(x) = 8000+20x. The marginal revenue function will be R’(x) = (-x/25) + 110. At the production level of 400, the marginal cost will stay at $20 while the marginal revenue will be $125. This shows …show more content…
The lowest amount of revenue I can possibly earn is 0, thus I will set R(x) to 0 to find the feasible range of x. The domain for this function will thus be 0≤ x ≤2750. Profit is calculated by subtracting revenue and total cost. The profit function will then be P(x) = (-x2/.04) +90x-8000. Break-even points appear when the profit equals to 0. Setting P(x) to be 0, the break-even points will be when 93 and 9860 chairs are sold (rounded to the nearest unit). This means that when either 2157or 51,240 phone cases are sold, there will be no profit as the revenue will equal the total cost. Amount of phone cases between those numbers sold will generate a …show more content…
The orange region on the left should actually by the first quadrant.
As my business obtain the equation which the price $p is given as a demand for x and x is the number of phone cases that retailer is likely to buy. To determine the feasible region, set the price-demand equation to zero. Which agency the smallest number of products that I can feasibly sell using f(p) is 0 and the largest number of products I could feasibly sell is 2,750 because the feasible range is 0≤p≤2,750 when p is greater than or equals 0. The revenue function for my phone cases is equal
The formula to calculate elasticity is(relative rate of change in demand)/(relative rate of change in price)=-(pf'(p))/(f(p))= . Using the price-demand equation, the function of elasticity will be E(p)= At the price $94, the elasticity will be 5.875. According to the definition, when the elasticity of demand is greater than 1, the demand is elastic. This means that if I increase my price, the increase in revenue caused by the price change will be smaller than the decrease in revenue due to the decrease in quantity demanded, causing a loss. While if I decrease my price, the increase in revenue because of more people wanting my products will exceed the loss caused by the reduction in price, leading to a

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