Paper 4: Fundamentals of Business Mathematics & Statistic

3.52 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Calculus

Hence the maximum profit is obtained by putting x = 3 in the revenue equation R = 15xe– x/3

= 15.3.e– 1 = =2.72^45 =16.54.

Example 110 :For a certain establishment, the total revenue function R and the total cost function C are

given by

R = 83 x – 4x^2 – 21 and c = x^3 – 12x^2 + 48x + 11 where x = output.

Obtain the out put for which profit is maximum.

Solution:

Profit (p)= Revenue – Cost

= 83x – 4x^2 – 21 – (x^3 – 12x^2 + 48x + 11) = – x^3 + 8x^2 + 35x – 32

For max. dpdx=0. i.e.,dxd (−x 8x 35x 32 0^3 +^2 + − )=

or, – 3x^2 + 16x + 35 = 0 or, 3x^2 – 16x – 35 = 0

or, (x – 7) (3x + 5) = 0 or, x = 7, – 5/3

Again ( )

(^22)
2
d p d 3x 16x 35 6x 16
dx =dx − + + = − +
For x
2
2
7,d p 6.7 16 42 16 26 0,
dx
= = − + = − + = − < maximum
∴ the profit is maximum at x = 7.
Example 111 :Find two positive numbers whose product is 64 having minimum sum.
Let the two positive numbers be x and y. By question xy = 64 or y =^64 x.
Let s be the sum of numbers of that s = x + y
or, s x= +^64 x. Now diff. w.r.t.x, we get, dxds 64= −1 .x 2 For maximum ddxs=^0
i.e.,^1 −^64 x 2 =^0 or, x^2 = 64 or x, = 8, –8
2
2 3
d s 0 2.64.
dx = + x At
2
2 3
x 8,d s 2.64 0
= dx 8= > , minimum
∴ s is minimum for x = 8, the other number y=64 64x 8= = 8
∴ reqd. positive numbersare 8 and 8.