Answer to Question #272572 in Calculus for sumu

Question

Nalini wants to open a small restaurant stand selling specialty burgers. The fixed

cost component on a daily basis is Rs 2500. For each burger made, the cost is Rs

50. In addition the daily special cost is 0.25x2 where x is the number of burgers she

will make for the day.

a) Write down the expression of the Total Cost function for a day (1 mark)

b) Write down the expression for the Average Cost function for a day

c) How many burgers should she make a day to minimize the cost per burger per

day? ( 4 marks)

d) What is the total cost per day associated with the minimized cost per burger per

day?

e) The price per burger is 250. What is the maximum daily revenue Nalini can earn

provided she decides to make at the minimum unit cost per burger?

f) What is the daily profit she would earn in e) ?

g) How many burgers would she have to sell to break even on a daily basis?

Answer 1

a)

$$TC(x)=2500+50x+0.25x^2, x\geq0$$

b)

$$Average\ Cost=\dfrac{TC}{x}=\dfrac{2500+50x+0.25x^2}{x}$$

$$=\dfrac{2500}{x}+50+0.25x, x>0$$

c)

$$(Average\ Cost)'_x=-\dfrac{2500}{x}+0.25$$

Find the critical number(s)

$$(Average\ Cost)'_x=0=>-\dfrac{2500}{x^2}+0.25=0$$

$$x^2=10000$$

$$x=\pm100$$

We consider that $x>0.$

If $0<x<100, (Average\ Cost)'_x<0,$

$Average\ Cost$ decreases.

If $x>100, (Average\ Cost)'_x>0,$

$Average\ Cost$ increases.

The cost per burger per day has the absolute minimum when she makes $100$ burgers a day.

d)

$$TC(100)=2500+50(100)+0.25(100)^2$$

$$=10,000$$

e)

$$R(100)=250(100)=25,000$$

f)

$$P(100)=R(100)-TC(100)$$

$$=25000-10000=15,000$$

g)

$$P(x)=0=TC(x)-R(x)$$

$$2500+50x+0.25x^2-250x=0$$

$$x^2-800x+10000=0$$

$$x^2-800x+160000=150000$$

$$(x-400)^2=150000$$

$$x=400\pm\sqrt{150000}$$

$$x=400\pm100\sqrt{15}$$

$$x=13\ or \ x=787$$