my challenge problem - - maximum product

[lookagain]

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\(\displaystyle Let \ \ x, \ y, \ \ and \ \ z \ \ be \ \ the \ \ variables.\)

\(\displaystyle Let \ \ a, \ b, \ c, \ \ and \ \ d \ \ be \ \ the \ \ variable \ \ constants.\)

\(\displaystyle Let \ \ a, \ b, \ c, \ d, \ x, \ y, \ \ and \ \ z \ \ belong \ \ to \ \ the \ \ set \ \ of \ \ the \ \ positive \ \ real \ \ numbers.\)

\(\displaystyle If \ \ ax + by + cz = d, \ \ then \ \ determine \ \ the \ \ maximum \ \ value \ \ of \ \ the \ \ product \) \(\displaystyle xyz \ \ in \ \ terms \ \ of \ \ a, \ b, \ c, \ \ and \ \ d.\)

 

[JeffM]

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\(\displaystyle Let \ \ x, \ y, \ \ and \ \ z \ \ be \ \ the \ \ variables.\)

\(\displaystyle Let \ \ a, \ b, \ c, \ \ and \ \ d \ \ be \ \ the \ \ variable \ \ constants.\)

\(\displaystyle Let \ \ a, \ b, \ c, \ d, \ x, \ y, \ \ and \ \ z \ \ belong \ \ to \ \ the \ \ set \ \ of \ \ the \ \ positive \ \ real \ \ numbers.\)

\(\displaystyle If \ \ ax + by + cz = d, \ \ then \ \ determine \ \ the \ \ maximum \ \ value \ \ of \ \ the \ \ product \ \ xyz \ \ in \ \ terms \ \ of \ \ a, \ b, \ c, \ \ and \ \ d.\)

Given: \(\displaystyle Given\ ax + by + cz = d,\ find\ maximum\ of\ xyz.\)

I am going to ignore the special cases where a = 0, b = 0, or c = 0. Highlight below to see my answer.

\(\displaystyle Let\ w = xyz + u(d - ax - by - cz).\)

\(\displaystyle \dfrac{\delta w}{\delta x} = yz - ua \implies \dfrac{\delta w}{\delta x} = 0\ \ iff\ \ yz = au \implies u = \dfrac{yz}{a}.\)

\(\displaystyle \dfrac{\delta w}{\delta y} = xz - ub \implies \dfrac{\delta w}{\delta y} = 0\ \ iff\ \ xz = bu = \dfrac{byz}{a} \implies x = \dfrac{by}{a}.\)

\(\displaystyle \dfrac{\delta w}{\delta z} = xy - uc \implies \dfrac{\delta w}{\delta y} = 0\ \ iff\ \ xy = cu \implies \dfrac{by^2}{a} = \dfrac{cyz}{a} \implies z = \dfrac{by}{c}.\)

\(\displaystyle \dfrac{\delta w}{\delta u} = (d - ax - by - cz) \implies \dfrac{\delta w}{\delta u} = 0\ \ iff\ \ d = ax + by + cz \implies \)

\(\displaystyle d = \dfrac{aby}{a} + by + \dfrac{cby}{c} = 3by \implies y = \dfrac{d}{3b} \implies\)

\(\displaystyle x = \dfrac{b}{a} * y = \dfrac{b}{a} * \dfrac{d}{3b} = \dfrac{d}{3a}\ \ and\ \ z = \dfrac{b}{c} * y = z = \dfrac{b}{c} * \dfrac{d}{3b} = \dfrac{d}{3c}.\)

\(\displaystyle Constrained\ maximum\ of\ xyz = \dfrac{d^3}{27abc}.\)