Math Writing Assignment 1

J&E Smartphone Company

            J&E Smartphone Company is a new company producing smartphone. This is a smartphone with multifunction and touchscreen. It was produced for the elder people because of the big screen and oversize font. It’s easy for them to read the message and watch video. It’s convenient for them to have facetime with their children far away from them. The variable cost of this smartphone is $90.  The fixed cost of the smartphone is $4000 per month, including rent and utilities for business. By using this fixed production costs and the product variable cost multiplied by the quantity (x), so the cost function is C(x) = $4000 + 90x. If J&E have production budget constraint, then I can compute the maximum number of units that can be produced per month. However, J&E do not have a production budget constraint. If there was a budget constraint, suppose it is $184,000, then maximum production capability would be 2000 units, that is calculated as $184000 = $4000 +90x, so x = 2000 units.

 

            Generally, revenue is calculated by the units sold(x) multiple price (p). x = f(p) = 12000-75p. So p = (12000-x)/75. Because x > 0, p > 0, 12000-75p > 0, 0 < p < 160, (12000-x)/75 > 0, so 0 < x < 12000. Then the revenue function is R(x) = 160x – x^2/75, which is from x*p = x*(12000-x)/75 = 160x – x^2/75. The profit is the target of a business company., so the profit function is P(x) = R(x) – C(x) = 160x – x^2/75 – (4000 + 90x), so P(x) = 70x – x^2/75 – 4000. Break-even point come up when my company doesn’t gain or loss money, it means that the total revenue equaling to the total cost, so R(x) = C(x), 160x – x^2/75 = 4000 + 90x, so x = 57.77 or x = 5192.22. I have rounded these values for break event to avoid partial products. So, my feasible range is (58, 5193). Maximum feasible is 58 units and Minimum feasible is 5193 units and also the breakeven points. When the sold units are 58 and 5193, the total revenue is same as the total cost. My company will have no profit or loss.

 

            At the production level of 3000 units, the total production cost is C(x) = 4000 + 90 * 3000 = $274000, total revenue is R(x) = 160 * 3000 – 3000^2/75 = $360000, total profit is P(x) = 70 * 3000 – 3000^2/75 – 4000 = $86000. During this month, we produced 3000 units of smartphone, which costs $274000, and we sold it out and earned $360000, deducing the cost, we made a profit of $86000. At the same production level, we solve the marginal cost C’(x), marginal revenue R’(x), and marginal profit P’(x) with first derivative of the cost function, revenue function and profit function and then substitute x with 3000, C’(x) = 90, C’(3000) = 90. At the production level of 3000 units, the cost will be increasing at the rate of $90 per unit.   R’(x) = 160 – 2x/75, R’(3000) = 80. At the production level of 3000 units, the revenue will be increasing at the rate of $80 per unit. P’(x) = 70 – 2x/75, P’(3000) = – 10. At the production level of 3000 units, the profit will be decreasing at the rate of $10 per unit.

E(p) = (–p * f’(p)) / f(p)

        = (–p) * (–75) / (12000 – 75p)

        = 75p/(12000-75p)

        = p / (160 – p).

When E(p) < 1, p / (160 – p) < 1, p < $80, the demand is inelastic. Demand is not sensitive to changes in price, that is, percentage change in price produces a smaller percentage change in demand. The revenue is increasing when a price increase;

when E(p) = 1, p = $80, the demand is unit elastic. A percentage change in price produces the same percentage change in demand. The revenue keeps the same whenever the price increase or decrease;

when E(p) >1, p > $80 the demand is elastic. Demand is sensitive to changes in price, that is, a percentage change in price produces a larger percentage change in demand. The revenue is decreasing when a price increase.

 

            In order to get the maximize profit, first, we should first derivative profit function P(x), P’(x) = 70 – 2x/75, when P’(x) = 0, we can get x = 2625, then we should second derivative, P’’(x) = – 2/75. When x = 2625, P’’(2625) = – 2/75, that means when x = 2625, we get the maximize profit equals to $87875, P(2625) = 70 * 2625 – 2625^2/75 – 4000 = $87875. At same time p(x) = (12000-x)/75, p(2625) = (12000-2625)/75 = $125 per unit.

We can also double check it:

P(x) = R(x) – C(x).

R(2625) = 160*2625 – 2625^2/75 = $328125,

C(2625) = 4000 + 90 * 2625 = $240250,

P(2625) = R(2625) – C(2625) = $328125 – $240250 = $87875.  

 

As a result, when price is $125 per unit, the production level of 2625 units, we get the maximize profit value $87875. So J&E company should decrease the production level from 3000 units to 2625 units to get more profit from $86000 to $87875.