How do I start this problem?

Question

find all real values for $x,y$ and $z$ such that

$x + y + z = 51$

and

$xyz=4000$

given that

$z\geq 25$ and $0<x\leq 10$

I believe that the only possible solution is $10, 16, 25$ but I do not know how to prove this. I have trying subbing in various ways but I don't think that has helped. The intermixing of inequalities and equalities is something I am not familiar with, so I do not know how to proceed. What kind of techniques are needed?

Any hints to solving this would be appreciated, as I would still like to try solve it myself. If I cannot solve it from those I will reply for more clarification.

Answer 1

The solution by Ahmad can be simplified.

Substituting $z=51-x-y\quad$ we get $\quad xyz=xy(51-x-y)=51xy-x^2y-xy^2=4000$

$y^2+(x-51)y+\frac{4000}x=0\qquad$

Assuming there are real solutions, then they are given by $2y=51-x\pm\sqrt{\Delta}\\$ $\text{where }\Delta=(x-51)^2-\frac{16000}x$

Edit: a side study shows $\Delta<0$ when $0<x<1$ so we can assume $x\ge 1$ from now on, it was necessary to justify the statement below (the last implication).

The condition that one of the solution is greater than $25$ implies:

$51-x\pm\sqrt{\Delta}\ge 50\implies \pm\sqrt{\Delta}\ge x-1\implies \Delta\ge (x-1)^2$

Substitution the value for $\Delta$ and using $a^2-b^2$ identity we get:

$(-50)(2x-52)\ge \frac {16000}x\iff 26-x\ge \frac{160}x\iff x^2-26x-160=(x-10)(x-16)\le 0$

Which happens when $x\in[10,16]$.

But since the problem states that $0<x\le 10$ then $x=10$

Now $y^2+(x-51)y+\frac{4000}x=y^2-41y+400=(y-16)(y-25)=0$

Gives the remaining solutions $y=16$ and $z=25$, and we check it verifies the initial problem.

All we had from now where mostly "$\implies$" statements, but it's enough to prove the solution is unique.

Answer 2

Since $0<x \leq 10$ and $z \geq25$ so $y>0$ because $x y z=4000$ and that can happens only if $y$ is positive number.

Now from the first equation we have that $x=51-y-z$.

Substitute that in the second equation to get $(51-y-z)y z=4000$ expanding the brackets and solving for $z$ we get that $z=\frac{-y^2+51y-\sqrt{\left(y^2-51 y\right)^2-16000 y}}{2 y}$ and $z=\frac{-y^2+51y+\sqrt{\left(y^2-51 y\right)^2-16000 y}}{2 y}$

Now we have that $z\geq 25$ so $\frac{-y^2+51y+\sqrt{\left(y^2-51 y\right)^2-16000 y}}{2 y}\geq 25$ or$\frac{-y^2+51y-\sqrt{\left(y^2-51 y\right)^2-16000 y}}{2 y}\geq 25$

The part $\frac{-y^2+51y-\sqrt{\left(y^2-51 y\right)^2-16000 y}}{2 y} \geq 25$ have no solution for $y$ , leave the square root part on left and all the rest to the right we get that $\frac{-\sqrt{\left(y^2-51 y\right)^2-16000 y}}{2 y} \geq -\frac{1}{2}+\frac{y}{2}$ obviously the left side is negative so when the right side is positive its false, and the right side is positive when $-\frac{1}{2} +\frac{y}{2} >0$ or when $y>1$.

What about $0<y<1$ ?, well the terms inside the square-root will be negative and thus will not have any meaning in the Real domain,(i leave this for you to prove, hint : $0<y^4<y^3<y^2<y<1$ ).

So we are left with $\frac{-y^2+51y+\sqrt{\left(y^2-51 y\right)^2-16000 y}}{2 y} \geq 25$ or $\frac{+\sqrt{\left(y^2-51 y\right)^2-16000 y}}{2 y} \geq -\frac{1}{2}+\frac{y}{2}$ and when $0<y<1$ the the inequality have no meaning in the real domain so $y>1$ (the same proof as above).

multiply by $2y$ and since $y>0$ this does not change the inequality and we get that $\sqrt{\left(y^2-51 y\right)^2-16000 y} \geq y^2-y$

square both sides (since they are both positive) that does not change the inequality to get $\left(y^2-51 y\right)^2-16000 y\geq (y^2-y)^2$ expanding the brackets we get that $y^4-102 y^3+2601 y^2-16000 y\geq y^4-2 y^3+y^2 $ take all terms to one side we get that $-100 y^3+2600 y^2-16000 y\geq 0$ or $-100y(160 - 26 y + y^2) \geq 0$ solving for $y$ (quadratic equation) we get that $10\leq y \leq16$ or $y\leq 0$ (which is false since $y>0$).

Now we need to make sure that $ 0<x =51-y-z \leq 10$

So $51-y-\frac{-y^2+51y+\sqrt{\left(y^2-51 y\right)^2-16000 y}}{2 y} \leq 10$ or $51-y-\frac{-y^2+51y-\sqrt{\left(y^2-51 y\right)^2-16000 y}}{2 y} \leq 10$

taking the same approach as above one will reach that $16 \leq y \leq 25$ or $y \leq 0$ and (since $y>0$ this is false ).

So in the end $y$ must be in the interval $[10,16]$ and $[16,25]$ and the only value in both intervals is $y=16$ and solving for $x,z$ gives that $x=10,z=25$.

Answer 3

Solving for $y$ gives $\displaystyle 51-x-z=\frac{4000}{xz}\;\;$, so $\;51xz-x^2z-xz^2=4000\;$ with $0<x\le10$ and $z\ge25$.

Let $f(x,z)=51xz-x^2z-xz^2$ on the rectangular region $0\le x\le10, \;25\le z\le51$ $\hspace{2 in}$(since $z>51\implies y<0\implies xyz<0$);

we will show that $f$ has its maximum value of $4000$ when $x=10, z=25$:

1) On the left side, $g(z)=f(0,z)=0$ for $25\le z\le 51$.

2) On the right side, $g(z)=f(10, z)=410z-10z^2$ is decreasing on $[25,51]$, so its maximum occurs $\hspace{3.4 in}$when $x=10$ and $z=25$.

3) On the top, $h(x)=f(x,51)=-51x^2$ has a maximum value of 0.

4) On the bottom, $h(x)=f(x, 25)=25(26x-x^2)$ is increasing on $[0,10]$, so its maximum occurs $\hspace{3.4 in}$when $x=10$ and $z=25$.

5) Solving $f_{x}=51z-2xz-z^2=0$ and $f_{z}=51x-x^2-2xz=0$ gives the critical points $\hspace{.12 in}(0,0), (51, 0), (0,51), (17, 17)$, and none of these points are in the interior of the region being considered.

Therefore $f(10,25)=4000$ is the maximum value of $f$ on the region $0\le x\le 10, 25\le z\le 51$;

so $x=10, z=25$, and $y=16$ is the only solution.