the sum of 2 numbers...minimize the sum of their cubes.help!

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xkouki

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Nov 12, 2005
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Okay we have a problem..
The question is, if you have two non-negative numbers and their sum is 12, what is the minimum value of the sum of their cubes?......

First i set it up as x + y = 12 ....
Then x^3 + y^3 = minimum......
so i minimized by finding the derivative, 3x^2 + 3y^2= 0 .....

but after that i draw a blank! I'm not sure where to go from here..please help!
 
Use the fact that the sum is twelve.

. . . . .first number: x
. . . . .second number: y
. . . . .sum: x + y = 12

. . . . .second number: y = 12 - x

. . . . .first cube: x<sup>3</sup>
. . . . .second cube: (12 - x)<sup>3</sup>
. . . . .sum: x<sup>3</sup> + (12 - x)<sup>3</sup> = f(x)

Minimize.

Eliz.

Edit: Correcting typo in exponent.
 
my answer came out to be x=3 and x=-4, since the numbers are non-negative that leaves me with only x=3, so i plugged it back in..and i got the minimum value to be 6642....it seems rather too large?!?!....

(x=3, y=9) ....x^3 + y^3= 6642.........do you think i made a mistake?.
Thanks so much for ur help!!!!!!
 
xkouki said:
my answer came out to be x=3 and x=-4, since the numbers are non-negative that leaves me with only x=3, so i plugged it back in..and i got the minimum value to be 6642....it seems rather too large?!?!....

(x=3, y=9) ....x^3 + y^3= 6642.........do you think i made a mistake?.
Click to expand...
You probably made a mistake determining the derivative of f(x). You should get:
df/dx = 3x^2 - 3(12 - x)^2
Now set (df/dx)=0 and solve for x. You should get x=6 and y=(12-x)=6 for the answer.

.
 
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