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# Find two numbers whose difference is 120 and whose product is a minimum.

**Solution:**

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

x - y = 120 --- (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 - 120

dy/dx = dx/dx - d(120)/dx

Since 120 is constant d(120)/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 = 120

(-x) - x = 120

-2x = 120

x = -60

And hence

y = 60

Also

d²P/dx² = dy/dx + dx/dx = 1 + 1 = 2

Since second derivative is +ve

P = xy is minimum when x = -60 and y = 60

## Find two numbers whose difference is 120 and whose product is a minimum.

**Summary:**

The two numbers whose difference is 120 and whose product is a minimum are 60 and -60.

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