Find the smallest and highest value of the product $xyz$

Question

Find the smallest and highest value of the product $xyz$ assuming that:

$x + y + z = 10$ and $x^2 + y^2 + z^2 = 36$.

I calculated this:

$x+y+z=10 => (x+y+z)^2=10^2$

$x^2+y^2+z^2+2xy+2yz+2zx=100$

$(x^2+y^2+z^2+2xy+2yz+2zx)-(x^2+y^2+z^2)=100-36$

$2xy+2yz+2zx=64$

$xy+yz+zx=32$

I'm stuck. What is the next step to this exercise?

My idea is to show the equation using one variable and after computing the derivative reach global extremes.

Answer 1

We have $$x^2+y^2=36-z^2$$ and $$x+y=10-z,$$ which gives $$(10-z)^2-2xy=36-z^2$$ or $$xy=32-10z+z^2$$ and $$xyz=32z-10z^2+z^3.$$

Also, $$(x+y)^2\geq4xy,$$ which gives $$3z^2-20z+28\leq0$$ or $$2\leq z\leq\frac{14}{3}.$$ Can you end it now?

I got $\max(xyz)=\frac{896}{27}$ and $\min(xyz)=32$.

Answer 2

It seems like a better start if you eliminate a variable (or two...) Thus $z= 10-x-y$, so $x^2+y^2+100-2(x+y)+(x+y)^2=36$, that is, $x^2+y^2+xy-x-y+64=0$. You are looking for the maximum of $xy(10-x-y)$.

If you do not want to look for anything creative, Lagrange multiplier surely works.

Answer 3

Use the the method of Lagrange multipliers. The problem is $$ \left\{ \begin{array}{rl} \min & x\cdot y\cdot z \\ \mbox{such that} & x+y+z=10\\ & x^2 + y^2 + z^2 = 36 \end{array} \right. $$ By Lagrange multiplier theorem the critical points of $f (x, y, z)=x\cdot y\cdot z$ satisfying the constraints $g(x,y,z)= x+y+z-10=0$ and $h(x,y,z)=x^2+y^2+z^2-36=0$ is solution of $$ \left\{ \begin{array}{rl} \frac{\partial}{\partial x}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\ \frac{\partial}{\partial y}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\ \frac{\partial}{\partial z}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\ \frac{\partial}{\partial \lambda}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\ \frac{\partial}{\partial \mu}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\ \end{array} \right. $$ Here, $ \mathcal{L}(x,y,z,\lambda,\mu)= f(x,y,z)-\lambda g(x,y,z)-\mu h(x,y,z) $ In this case, $$ \left\{ \begin{array}{rlcl} y\cdot z-\lambda -2x\mu=&0 &\hspace{1cm} & (1)\\ x\cdot z-\lambda -2y\mu=&0 &\hspace{1cm} & (2)\\ x\cdot y-\lambda -2z\mu=&0 &\hspace{1cm} & (3)\\ x+y+z=&10 &\hspace{1cm} & (4)\\ x^2+y^2+z^2=&36 &\hspace{1cm} & (5)\\ \end{array} \right. $$

Answer 4

Consider the equation: $A^3-10A^2+32A-xyz=0$, whose roots are $x,y,z$. We need to find the maximum and minimum values of: $$xyz=A^3-10A^2+32A=f(A).$$

FOC: $$f'(A)=3A^2-20A+32=0 \Rightarrow A_1=\frac83; \ \ A_2=4.$$ SOC: $$\begin{align}f''(A)&=6A-20; \\ f''\left(\frac83\right)&=-4<0 \Rightarrow f\left(\frac83\right)=\frac{896}{27} \ (\text{max});\\ f''(4)&=4>0 \Rightarrow f(4)=32 \ (\text{min}).\end{align} $$

Answer 5

As an alternative, we can also try to solve the problem by a geometrical intepretation, indeed the constraints represent a circle which is the intersection between

• the plane $x+y+z=10$
• the sphere $x^2+y^2+z^2=36$

therefore the center of the circle is located at

$$OC=\left(\frac{10}{3},\frac{10}{3},\frac{10}{3}\right)$$

and the radius of the circle is

$$R^2=36-|OC|^2 \implies R^2=\frac83 \implies R=\frac{2\sqrt 6}{3}$$

To parametrize the circle let consider at first the circle in $x-y$ plane and centered in the origin

$$x^2+y^2=\frac83 \iff \left(\frac{2\sqrt 6}{3}\cos \alpha, \frac{2\sqrt 6}{3} \sin \alpha,0\right)$$

and since the rotation matrix around $u=\left(\frac{\sqrt 2}{2},-\frac{\sqrt 2}{2},0\right)$ of an angle $\theta=\arccos \left(\frac{\sqrt 3}{3}\right)$ is given by

$$M = \begin{bmatrix} \frac12+\frac{\sqrt 3}{6} & -\frac12 +\frac{\sqrt 3}{6} & \frac{\sqrt 3}{3} \\ -\frac12 +\frac{\sqrt 3}{6} & \frac12+\frac{\sqrt 3}{6} & \frac{\sqrt 3}{3} \\ -\frac{\sqrt 3}{3} & -\frac{\sqrt 3}{3} & \frac{\sqrt 3}{3} \end{bmatrix}$$

the intersection circle can be parametrized as follow

• $x(\alpha)=\frac{10}{3}+\frac{2\sqrt 6}{3}\left(\frac12+\frac{\sqrt 3}{6}\right)\cos \alpha +\frac{2\sqrt 6}{3}\left(-\frac12 +\frac{\sqrt 3}{6}\right)\sin \alpha =\frac{\sqrt 6}{3}\left[\frac{5\sqrt 6}3+\left(1+\frac{\sqrt 3}{3}\right)\cos \alpha +\left(-1+\frac{\sqrt 3}{3}\right)\sin \alpha\right]$

• $y(\alpha)=\frac{10}{3}+\frac{2\sqrt 6}{3}\left(-\frac12 +\frac{\sqrt 6}{6} \right)\cos \alpha +\frac{2\sqrt 6}{3}\left(\frac12+\frac{\sqrt 6}{6}\right)\sin \alpha =\frac{\sqrt 6}{3}\left[\frac{5\sqrt 6}3+\left(-1+\frac{\sqrt 3}{3}\right)\cos \alpha +\left(1+\frac{\sqrt 3}{3}\right)\sin \alpha\right]$

• $z(\alpha)=\frac{10}{3}-\frac{2\sqrt 2}{3}\left(\cos \alpha +\sin \alpha \right)=\frac{2\sqrt 2}{3}\left(\frac{5\sqrt 2}{2}-\cos \alpha -\sin \alpha \right)$

and finally we obtain

$$xyz(\alpha)=\frac{8}{27}\left(\sqrt 2 \cos (3\alpha )-\sqrt 2 \sin (3\alpha)+110\right) \quad \alpha \in [0,2\pi)$$

that is

$$xyz(\alpha)=\frac{8}{27}\left[ 2 \sin \left(\frac{\pi}4-3\alpha\right)+110\right] \quad \alpha \in [0,2\pi)$$

and therefore

$$32\le xyz \le \frac{896}{27}$$

form the plot of we can see that the maximum and the minimum values are reached each one in threee different points of the domain.