Question

Find three positive numbers, whose sum is 30 and whose product is maximum.​

Answer 1

10 +10+10 =30

and the product is 10×10×10

which gives

=1000

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Answer 2

By the AM-GM inequality,

30=x+z+y2+y2≥4⋅(xy2z4)14⟹xy2z≤4⋅(152)4

The equality holds when x=y2=z=152.