The problem involves finding the maximum and minimum values of the expression \(x + y + z\) given the equation \(4^{\sqrt{5x + 9y + 4z}} - 68 \times 2^{\sqrt{5x + 9y + 4z}} + 256 = 0\). By substituting \(\lambda = 2^{\sqrt{5x + 9y + 4z}}\), the quadratic equation \(\lambda^2 - 68\lambda + 256 = 0\) yields solutions \(\lambda = 4\) and \(\lambda = 64\). This leads to two cases for \(5x + 9y + 4z\): when it equals 4, the minimum value of \(x + y + z\) is \(4/9\); when it equals 36, the maximum value is 9. Consequently, the product of the minimum and maximum values is \(4\).
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Opalg said:[sp]Let $\lambda = 2^\sqrt{5x+9y+4z}$. Then the equation becomes $\lambda^2 - 68 \lambda + 256 = 0$, with solutions $\lambda = 4 = 2^2$ and $\lambda = 64 = 2^6.$ Thus $2^\sqrt{5x+9y+4z} = 2^2$ or $2^6$, and so $5x+9y+4z = 4$ or $36$. When $5x+9y+4z = 4$ we get $(x+y+z)_{\min} = 4/9$ by taking $(x,y,z) = (0,4/9,0)$. When $5x+9y+4z = 36$ we get $(x+y+z)_{\max} = 9$ by taking $(x,y,z) = (0,0,9)$. Therefore $(x+y+z)_{\min}\times (x+y+z)_{\max} = 4$.[/sp]