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

[juandiaz]

Find maximum value of [imath]xy(72 - 3x - 4y)[/imath], [imath]x > 0,\; y > 0[/imath]

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

 

[blamocur]

Is this a homework problem? If it is, which topic is it related to?

 

[juandiaz]

This is my work now but i’m not sure.

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

[imath]\qquad a = 3x[/imath]
[imath]\qquad b = 4y[/imath]
[imath]\qquad c = 72 - 3x - 4y[/imath]

Then:

[imath]\qquad \dfrac{a + b + c}{3} = \dfrac{72}{3} = 24[/imath]

So:

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

[imath]\qquad \sqrt[3]{(3x)(4y)(72 - 3x - 4y)\;} \le 24[/imath]

[imath]\qquad 12xy(72 - 3x - 4y) \le 24 \cdot 24 \cdot 24[/imath]

[imath]\qquad xy(72 - 3x - 4y) \le \dfrac{24 \cdot 24 \cdot 24}{12} = 1152[/imath]

Maximum value is [imath]1152[/imath]

 

[blamocur]

This is my work now but i’m not sure.View attachment 36101

Neat! I get the same answer by different methods.

 

[khansaheb]

Neat! I get the same answer by different methods.

The other method could be:

https://tutorial.math.lamar.edu/classes/calciii/relativeextrema.aspx