Maths question for your help on Cost and Revenue Functions

Question

Suppose the demand function is q^d= 1.5p^2 – 30p + 50 and supply function is q^s =0.25p^2-50. Determine the price and quantity at which the market reaches its equilibrium point, if the prince domain is RM50<p< RM50

b)  a manufacturer has affixed cost of RM84,00 and a variable cost of RM 1.60 per unit made and sold. Selling price is RM 2.20 per unit.

i)  Find the cost, revenue and profit functions using Q for number of units

ii)  Find the break-even quantity; and

iii) What happens to the break-even quantity if the selling price rises to RM2.40 per unit?

c) Company TRQ Bhd has found that its demand function for its product is p=2320- 8q, where p represent the unit price and q is the quantity demanded for the product.

                i) Determiner the revenue function;

ii) Determiner the quantity that will maximize the total revenue; and

iii) Determine the maximum value of the revenue.

Answer 1

a) The equilibrium point is when supply and demand balance, so 1.5p^2-30p+50=0.25p^2-50. So 1.25p^2-30p+100=0. Multiply through by 4: 5p^2-120p+400=p^2-24p+80=0=(p-4)(p-20)=0. So p=4 or 20, both of which are less than RM50. I suspect that RM20 is the correct price, even though the domain given contains an error.

b)

i) Cost, c=84+1.60N; selling price, s=2.20N, where N is the number of items made and sold. Profit, p=s-c=2.20N-84-1.60N=0.60N-84. 

ii) Break-even quantity is N where c=s or p=0: 0.60N=84, N=84/0.60=140. 

iii) p=2.40N-84-1.60N=0.80N-84, so when p=0 N=84/0.80=105. The break-even quantity has reduced by 35 from 140 to 105.

c)

i) Revenue, r, from sales=pq. Substitute p=2320-8q: r=q(2320-8q)=2320q-8q^2=8(290q-q^2)=-8(q^2-290q+145^2)+8*145^2=-8(q-145)^2+8*145^2, (ii) which has a maximum value when q=145, because the first term becomes zero, iii) maximum r=RM168200 when q=145 (and p=RM1160).