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I am reviewing for my exam next week and in my notes we have this problem:

Given x,y,z as real numbers and $4^{\sqrt{5x+9y+4z}}-68\cdot 2^{\sqrt{5x+9y+4z}}+256=0$. Find the product of the minimum and maximum values of $x+y+z$.

The solution is as follows:

The given equation is equivalent to $u^2-68u+256=0$ with $u=2^\sqrt{5x+9y+4z}$. The solution of which is $u=2^2$ or $u=2^6$ which imply the following cases.

Case 1: When $5x+9y+4z=2^2$, $4(x+y+z)\leq 4\leq 9(x+y+z)$ so that $\frac{4}{9}\leq x+y+z\leq 1$. Case 2: When $5x+9y+4z=6^2$, we have $4\leq x+y+z\leq 9$.

Hence, $\frac{4}{9}\leq x+y+z \leq 9$ and the required product is $4$.

Now my question is how were the inequalities in the 2 cases above were established? I'm confused, thanks for any help.

james25
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1 Answers1

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Following your comment I'm adding the condition that $x$, $y$, $z$ are nonnegative.

You found that $5x +9y +4z\in\{4,36\}$ is necessary and sufficient.

Now write $$s:=x+y+z={1\over9}(5x+9y+4z)+{4\over9}x+{5\over9} z\geq {1\over9}(5x+9y+4z)\geq{4\over9}$$ with equality iff $x=z=0$ and $5x+9y+4z=4$, whence $y={4\over9}$. It follows that $s_\min={4\over9}$.

Similarly $$s={1\over4}(5x+9y+4z)-{1\over4}x-{5\over4}y\leq{1\over4}(5x+9y+4z)\leq9$$ with equality iff $x=y=0$ and $5x+9y+4z=36$, whence $z=9$. It follows that $s_\max=9$.

Therefore we may conclude that $s_\min\>s_\max =4$.

Christian Blatter
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