First slide

Application of diffrentiation

Question

Divide a number 100 into two parts such that their product is maximum.

Moderate

A

50,50
B

25,75
C

60,40
D

90,10

Solution

Let the two parts be x and (100-x)
Product,
$P = x (100 - x) = 100 x - x^{2}$
For P to be maximum or minimum,
$\frac{d P}{d x} = 100 - 2 x = 0$
x=50
$\frac{d^{2} P}{d x^{2}} = - 2$
i.e. P is maximum at x =50
Dividing equally the two parts are (50,50)

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