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akotoh · Oct 5, 2009, 08:27 AM

Please help me with this one...
Find two positive numbers whose product is 64, and whose sum is a minimum. Thanks!

Perito · Oct 5, 2009, 08:56 AM

X = first number
Y = second number

XY = 64
X + Y = minimum.

Substituting

X + 64/X = minimum

To find the minimum (or maximum), differentiate with respect to X and set the result to zero.

$\frac {d}{dX}(X + \frac {64}{X}) = 1 - \frac {64}{X^2} = 0$

$\frac {64}{X^2} = 1$

$64 = X^2$

$X = \pm\,8$

$Y = \frac {64}{X}$

Therefore

$Y = \pm \, 8$

If you want the sum to be a minimum then you must select X=-8, Y=-8

p.s.
... But, since the problem specified positive numbers, you select

X=8 and Y=8 :o

ebaines · Oct 5, 2009, 08:56 AM

I suggest you write out all the ways you can multiply two positive numbers to get 64, and see which has the smallest sum:

1*64
2*32
4*16
etc.

EDIT - I assumed (perhaps incorrectly) that the OP meant two positive "integers," in which case this technique is adequate. I also assumed (again, I suspect incorrectly) that this was a question for someone in elementary school, and hence the use of calculus as Perito suggested would be beyond the level of understanding of the OP. However, I see from some earlier posts that the OP is indeed studying derivatives, so Perito's technique is appropriate.

Unknown008 · Oct 5, 2009, 09:32 AM

Well, Perito you missed the part where it said that 'two positive numbers' :rolleyes:

Perito · Oct 5, 2009, 02:02 PM

Oops.