Math Writing Assignment 1

Excel Leather Shoes Company

Excel Leather Shoes Company or ELS is my company and it manufactures and sells leather shoes. The leather shoes are differentiated in the kind of shapes, and outlook and this enables to meet the needs and preferences of different customers .The variable cost per pair of the leather shoe is $30. This variable cost is computed by averaging the variable costs of a number of various kinds and rounded to the nearest dollar for the simplicity. These varieties that I sell have elegant shape and are highly recommendable outfit as an official wear. SLE has fixed production costs of $4000 per month. By using this fixed production costs and the product variable cost multiplied by the quantity (x), I derive my cost function as C(x) = 4000+30x. If I have production budget constraint, then I can compute the maximum number of units that can be produced per month. ELS do not have a production budget constraint. If there was a budget constraint, let say like $160,000, then maximum production capability would be 5200units, that is calculated as $16000= $4000+30x, where x= 5200 units.

In particular, revenue is calculated by multiplying the number of product units sold and the price, that R(x) = x*[p(x)] .Given that the price demand equation was x= f (p) = 12,000 -75p, then the revenue function would be R(x) = (12,000x –x^2)/75.This is computed as follows;

First making p the subject of formula of equation x= 12,000 -75p, we have p= (12000-x)/75, then multiplying x by p, to get revenue .Thus, R(x)=x*p =x*((12000-x)/75)= (12,000x –x^2)/75.

From this revenue(R(x)) and cost function (C(x)), I deduced the profit function, P(x) = – (x^2)/75 + 130x -4000, , that is calculated as illustrated ; P(x) = (160x –(x^2)/75) – (4000+30x) = – (x^2)/75 + 130x -4000

At breakeven points, my company would neither make profit nor loss and so, the revenue would be equal to the costs incurred. So, setting C(x) =R(x) or equating the profit function to zero. I deduced the break even points, and I have two values where the costs would equally be the same as the revenue generated. Notably, there is no restriction on the budget production, and consequently this feasible range would be the two breakeven points. The break even points are 31 and 9720 pairs. I have rounded these values for break event to avoid partial products. So, my feasible range is (31, 9720). The computations for getting break even points are illustrated as follows;

C(x) = R(x) = 4000+30x = (12,000x –x^2)/75),

Thus 4000-130x+ (x^2)/75=0, Either x= 30.86695062986564 or 9719.133049370135

Maximum feasible is 31 pairs and Minimum feasible is 9720 pairs and also the breakeven points.

At the production level of 3000 units , the total production cost would $16,000 and the revenue would be $360,000.Also, at the same production level , the marginal cost , marginal revenue and marginal profit would $30,$80, and $50 respectively .To get the values for the marginal cost , marginal revenue and marginal profit , I used the first derivative of the cost function , revenue function and profit function respectively .Then resulting expressions of marginal revenue and marginal profit ,I substituted x with 3000.This simply means that producing one more pair of the leather shoes would increase my costs by $30 , raise my revenue by $80 and also lead to an increase of my profit by $50.At the production of 3000 pairs per month , I would both increase my production and profits.

Computations for Marginal cost, Marginal revenue, and Marginal Profit at production of 3000

Marginal cost; C(x) = $4000+ 30x, C’(x) =30, C’ (1000) =30.

Marginal revenue; R(x) = (12,000x –x^2)/75, R’(x) = (12,000 –2x)/75),

R’ (3,000) = 160 – ((2*3000)/75) = 80.

Marginal Profit; P(x) = – (x^2)/75 + 130x - 4000, P’(x) = - 2x/75 + 130 ,

P(3000) = $50

At the optimal production, 4875 units are produced .I found this getting the first derivative of profit function, P(x) = – (x^2)/75 + (160- cost) x - 4000, and equating it to zero. The derivative of profit function is P’(x) = - 2x/75 + (160-cost). At this optimal production, the total costs, revenues, and profits at this point are $150,250, $463,125 and $312,875. These values are calculated by substituting x with 4875 units in the respective functions .The price that maximizes the profit is $ 95, computed by substituting this optimal production units in the demand equation ,x= f(p) = 12,000 -75p, .That is 4875 = 12,000 -75p, p =$ 95.

Elastic function E(p)= (-p * f’(p)) / f(p).  E(p) = 75p/(12,000-75p).

Inelastic: E(p)<1, 75p/(12,000-75p) <1, p<80. So 0<p<80 is Inelastic. A price increase will increase the revenue.

Unit elastic: E(p)=1, 75p/(12,000-75p) =1, p=80. So when p=80, it is unit elastic.

Elastic: E(p)>1, p>80. So 80<p<160, it is elastic. A price increase will decrease the revenue.

Like, any other business organization, the main objective is to maximize profits. Conclusively, marginal analysis helps managers to make sound decisions about the price and the production level that optimizes the company’s profit. In fact marginal analysis helps me in investigating the production of the next item through use of cost function and the benefit of an additional product.