Find two numbers whose difference is 100 and whose product is a minimum.

Solution:

Let one number be x and the second number be y.

x - y = 100 --- (1)

The product of two numbers is minimum therefore

P = xy is minimum which means

dP/dx = 0 and 

d²P/dx² should be +ve 

Rewriting (1) as below we have:

y = x - 100

dy/dx = dx/dx - d(100)/dx

Since 100 is constant d(100)/dx = 0

dy/dx = 1

Now,

dP/dx = ydx/dx + xdy/dx = 0 --- (2)

Since dy/dx = 1

dP/dx = y + x = 0

y = - x

Since 

y - x = 100

(-x) - x = 100 

-2x = 100

x= -50

And hence

y = 50

Also

d²P/dx² = dy/dx + dx/dx =1 + 1 = 2

Since second derivative is +ve

P = xy is minimum when x = -50 and y = 50

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Find two numbers whose difference is 100 and whose product is a minimum.

Summary:

The two numbers whose difference is 120 and whose product is a minimum are -50 and 50.