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20 12 Beijing mathematics college entrance examination questions liberal arts questions 17 how to find the maximum value of the last question? Solve! !
a^2+b^2+c^2=(600-b-c)^2+b^2+c^2=(b+c)^2- 1200(b+c)+600^2+b^2+c^2=2(b+c-300)^2-2bc+2*300^2

From a+b+c=600, A, B and C are all greater than or equal to 0, and B+C-300 =

So we get 2 (b+c-300) 2-2bc+2 * 300 2 =

That's 360000 at most.

A+b+c=600, and A, B and C are all greater than or equal to 0, then the sum of two numbers in A, B and C must be greater than or equal to 300, assuming B+C >; =300,

When the maximum value of a 2+b 2+c 2 is 360000.

-bc is the maximum value, that is, -bc=0, so b=0 or C = 0;;

B+c-300 is the maximum value, that is, b+c-300=300, that is, b+c=600, so a=0.

That is, a=0, b=0, c = 600 or a=0, b=600, c=0.