Question

The sum of two numbers is 12 find the minimum value of the sum of their cubes

a. 432
b. 346
c. 644
d. 244

Answer 1

5 and 7 is near lol =244

Answer 2

3 and 9 is near to =243 <smaller 1

Answer 3

x+y = 12

p(x)= x^3 + y^3 = (x+y)^3 - 3xy(x+y)

p(x) =12^3 -36xy
p(x)=1728 -36xy

minimise p(x) like we did last time set p'(x) = 0

Answer 4

first substitute y = 12 -x

Answer 5

the key to making this problem easier was the algebraic identity

(x+y)^3 -3xy(x+y) = x^3 + y^3

Answer 6

yup i heard early its positive but if the answer is true to be negative it must.

Answer 7

oooh er i think i made a mistake somewhere..

Answer 8

p(x)=1728 -36xy

p(x)=1728 -36x(12 - x)
= 1728 - 432x + 36x^2
p'(x) = 72x - 432

set p'(x) = 0

x = 432/72 = 6

y = 12-6 = 6

Answer 9

so 6^3 + 6^3 = 432

Answer 10

the answer is either 6,6 or 1,11 isn't it?
boundary problem, special values...
(6,6) = 432 (1,11) = 1,331
so I will go with 432

Answer 11

and I was wrong, the answer was going to be either 6,6 or 0,12

Answer 12

i solved it using a bit of calculus

Answer 13

eigen if this question is question again but the numbers are different how could i get like 12^3? in fastest way

Answer 14

thanks again

Answer 15

i get it

Answer 16

\[(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3\]

rearrange to get \[x^3 + y^3\]

\[x^3 + y^3 = (x+y)^3 - 3x^2y - 3xy^2 = (x+y)^3 - 3xy(x + y)\]

this is useful because we know what x+y is equal to :)

Answer 17

thanks

Answer 18

no problem, happy to help