Question

3. The revenue derived from the sale of shirts is represented by ? = 521? − ?2 where n is the number of shirts sold and R is the daily revenue. It is also known that the fixed cost is $5,364 per day plus a variable cost of $205 per shirt.

1. (a) Construct an equation to express the total cost per day (TC), in $, in terms of n.

2. (b) Find the profit function using the equation: Profit (P) = Revenue – Total Cost.

3. (c) Hence, determine the values of n for which the shop can make a profit.

4. (d) Find the maximum profit with the profit function found in (b) and the corresponding number of

shirts sold per day.

Answer 1

(a) Variable cost= Number of shirts*Variable cost per shirt

=n*205

Total cost= Fixed cost +Variable cost

=5,364+205n

(b)  Profit (P) = Revenue – Total Cost

=(521n-n^2)-(5,364+205n)

=521n-n^2-5,364-205n

=316n-n^2-5,364

(c) For break even (No profit situation), TC=Revenue

5,364+205n=521n-n^2

n^2+205n-521n+5,364=0

n^2 -316n+ 5,364=0

n^2-298n-18n+5,364=0

n(n-298)-18(n-298)=0

(n-18)(n-298)=0

So, n-18=0 or n-298=0TC=

n=18 or 298

When n=18,

TC=5,364+205(18)

=$9,054

Revenue=521(18)-(18)^2

=9378-324

=$9,054

When n=298,

TC=5,364+205(298)

=$66,454

Revenue=521(298)-(298)^2

=155,258-88,804

=$66,454

The firm will earn profit when n is greater than 18 and less than 298.

(d) For maximum profit,

Profit=316n-n^2-5,364

= -(n^2-316n+5,364)

=-(n^2-316n+24,964-24,964+5,364)

=-(n^2-316n+24,964-19,600)

=-(n^2-316n+24,964)+19,600

=-(n-158)^2+19,600

So, the maximum profit will be $19,600.

Number of shirts to get the maximum profit=158