MATH SOLVE

10 months ago

Q:
# Find two numbers whose difference is 68 and whose product is a minimum. (smaller number) (larger number)

Accepted Solution

A:

Let P be the product, x be the first number and y be the second number

therefore:

x-y=68........i

P=xy....ii

from i

x=68+y

hence

P=y(68+y)=y^2+8y

P=y^2+8y

The minimum product will occur at point where:

dP/dy=0

thus

dP/dy=2y+8=0

thus

y=-4

thus the value of x will be:

x=68-4=64

thus the product will be:

-4*64=-256 sq. units

therefore:

x-y=68........i

P=xy....ii

from i

x=68+y

hence

P=y(68+y)=y^2+8y

P=y^2+8y

The minimum product will occur at point where:

dP/dy=0

thus

dP/dy=2y+8=0

thus

y=-4

thus the value of x will be:

x=68-4=64

thus the product will be:

-4*64=-256 sq. units