Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 3.51

Solution:
Profit (P)= Sale – total cost = x × p – y
= x (300 – 2x) – (2x + 1000) = 298x – 2x^2 – 1000

For maximum profits, ddxP=0 i.e., 298 4x 0− =

or, x = 74.5 = 74 (as the number cannot be fraction and also x ≠ 75 as x ò 74.5)

Again

2

2

d P 4 0.
dx = − < Hence the profit will be maximum for 74 items.
Alternative way :

For profit maximisation we know MR = MC i.e ( ) ( )
d TR d TC
dx dx= ... (i)
TR = px = (300 – 2x) x = 300x – 2x^2

d TR( ) d(300x 2x 300 4x (^2) )
dx dx= − = −
Again TC = 2x + 1000, d TC(dx) = 2. Now by (i), we get
300 – 4x = 2 or, 4x = 298 or, x = 74.5 = 74.
(as the number cannot be fraction and also x ≠ 75 as x ò 74. 5)
Example 109 :The demand function for a particular commodity is y = 15e–x/3 for 0 ≤ x ≤ 8 where y is the price
per unit and x is the number of units demanted. Determine the price and the quantity for which the revenue
is maximum.
Revenue (R) = xy = x. 15e–x/3 = 15x. e–x/3
For maximum x/3 x/3 x/3 x/3
dR 15 x.e (^1) e 5xe 15e
dx 3

=^ − − + −^ = − − + −

For dRdx=0, we get – 5xe–x/3 + 15e –x/3 = 0

or, 5e–x/3 (3 – x) = 0, either e-–x/3 = 0, or, 3 – x = 0

i.e., x = ∞ (absurd) or x = 3

Also

(^2) x/3 x/3 x/3
2
d R 5 x.e. (^1) e 15e. 1
dx 3 3

= − − −^ + −^ + −^ −^

For

2

2

x 3,d R 0,

= dx < it is maximum.