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