posted by Anonymous .
Find two positve real numbers whose product is a maximum and whose sum of the first number and four times the second number is 120.
This is as far as I've gotten in solving it.
*first number is x.
*second number is y.
*xy=maximum, which is unknown.
*x+4y=120. Solve for x, which comes out to be x=-4y+120.
*Substitute -4y+120 for x in the equation xy, then solve for y. So (-4y+120)(y) = -4y(y-30) = y=0 and y=30.
Please show me what to do from here.
You know that 4(30)=120, so that would mean x would have to be 0, since x+4y=120. However, the problem states that both x and y must be real numbers. Since 2(2)=4, you need to find a number in which 2(2)(30)=120. What number times 2 will give 30? 15. x+4(15)=120. Solve for x. x=60. Thus, x+4y=120 is 60+4y=120, solving for y, you get y=15. So the first number is 60 and the second number is 15.