Find maximum value of xy(72 - 3x - 4y), x > 0, y > 0

juandiaz

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Find maximum value of xy(723x4y)xy(72 - 3x - 4y), x>0,  y>0x > 0,\; y > 0

I think of AM, GM inequality but don't know how.
 

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This is my work now but i’m not sure.

From a+b+c3abc  3\dfrac{a + b + c}{3} \ge \sqrt[3]{a\cdot b\cdot c\;}, let:

a=3x\qquad a = 3x
b=4y\qquad b = 4y
c=723x4y\qquad c = 72 - 3x - 4y

Then:

a+b+c3=723=24\qquad \dfrac{a + b + c}{3} = \dfrac{72}{3} = 24

So:

24(3x)(4y)(723x4y)  3\qquad 24 \ge \sqrt[3]{(3x)(4y)(72 - 3x - 4y)\;}

(3x)(4y)(723x4y)  324\qquad \sqrt[3]{(3x)(4y)(72 - 3x - 4y)\;} \le 24

12xy(723x4y)242424\qquad 12xy(72 - 3x - 4y) \le 24 \cdot 24 \cdot 24

xy(723x4y)24242412=1152\qquad xy(72 - 3x - 4y) \le \dfrac{24 \cdot 24 \cdot 24}{12} = 1152

Maximum value is 11521152
 

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