Find the maximum value of $f(x,y,z)=(1+x)(1+y)(1+z)$ over the closed unit ball in $\mathbb{R}^3$

Question

I got this question on my homework in Calculus 3 class. This exercise is about Lagrange multiplications and I can't get around it. The closed unit ball is compacted and $f$ is $C^1$ hence by Weierstrass theorem $f$ gets a minimum and a maximum over the ball.

Consider the constraint $g(x,y,z)=x^2+y^2+z^2-K$ where $K\in[0,1]$. We get that $\nabla g=(2x,2y,2z)$ is linearly independent since there is only one constraint. Thus by the Lagrange multiplier theorem, we get that $\nabla f = \lambda \nabla g$. From this point forth it's just nasty algebra which leads to nowhere. Any help will be appreciated.

P.S. Sorry about the writting, I'm new around here.

Answer 1

The equation $\nabla f = \lambda \nabla g$ comes to: $$((1 + y)(1 + z), (1 + x)(1 + z), (1 + x)(1 + y)) = \lambda(2x, 2y, 2z),$$ i.e. \begin{cases} (1 + y)(1 + z) = 2 \lambda x &&&(1) \\ (1 + x)(1 + z) = 2 \lambda y &&&(2)\\ (1 + x)(1 + y) = 2 \lambda z &&&(3) \end{cases} Take, say, $(2) - (1)$. You get, $$(1 + x)(1 + z) - (1 + y)(1 + z) = 2\lambda y - 2\lambda x,$$ or equivalently, $$(x - y)(1 + z) = -2\lambda (x - y).$$ From here, there are two possibilities: $x = y$ (and both sides are $0$), or we divide both sides by $x - y$ to get $1 + z = -2\lambda$.

We can do similar analysis, subtracting $(2)$ and $(3)$, or $(1)$ and $(3)$. We get similar choices: $y = z$ or $1 +x = -2\lambda$, and $x = z$ or $1 +y = -2\lambda$.

Let's suppose that $z \neq -1 - 2\lambda$. If this is the case, then $x = y$. Note that, if $y = z$, say, then $y \neq -1 - 2\lambda$, and so we would have $x = z$, i.e. $x = y = z$. The same conclusion can be similarly reached if we assume $x = z$ instead. Either which way, $z \neq -1 - 2\lambda \implies x = y = z$.

Contrapositively, if we assume $x, y, z$ are not all the same, this implies $z = -1 - 2\lambda$. Similar arguments show that $y$ and $x$ would also be $-1 - 2\lambda$. This leads to a contradiction; in any case, we may conclude that $x = y = z$.

Now that we know $x, y, z$ are the same, let's change all of them to $x$ in $(1),(2),(3)$. We get: $$(1 + x)^2 = 2\lambda x \iff x^2 + 2(1 - \lambda)x + 1 = 0.$$ We also have the constraint that $g(x, y, z) = 0$, i.e. $$x^2 + y^2 + z^2 = 1.$$ Given $x = y = z$, we get $$x = \pm\frac{1}{\sqrt{3}}.$$ Substituting it into the equation of $x$ and $\lambda$, $$\frac{1}{3} \pm \frac{2(1 - \lambda)}{\sqrt{3}} + 1 = 0.$$ Simplifying, $$\frac{2(1 - \lambda)}{\sqrt{3}} = \mp \frac{4}{3},$$ and so $$\lambda = 1 \pm \frac{2\sqrt{3}}{3}.$$

Answer 2

Kind of simpler argument would be like this - it is easy to see that at the point where maximum is attained (say, $x_0, y_0, z_0$ ) coordinates are not negative. Also, the expression is symmetric so we can assume that $x_0 \ge y_0 \ge z_0$. Say, we have $x_0 > y_0$. We want to find an $\epsilon >0$ such that the point $(x_0-\epsilon,y_0 + \epsilon,z_0)$ is in the unit ball. This is enough since if we have two distinct positive numbers $a,b, a>b$ then one has
$(a-\epsilon)(b+\epsilon) > ab$ if $\epsilon$ is small enough.
We want to check that $(x_0-\epsilon)^2 + (y_0 + \epsilon)^2 +z_0^2 \le 1$ After some simple algebra and using the fact that $x_0^2 + y_0^2 +z_0^2 = 1$ we get condition on $\epsilon$ i.e. $\epsilon \le x-y$. So we must have $x_0 = y_0$. The same argument can be repeated to show that $y_0=z_0$ Then using the constrain once again we get that at maximum $x_0 = y_0 = z_0 = \frac{1}{\sqrt3}$

Answer 3

No Lagrange multipliers are necessary to answer this question. The inequality between the geometric and arithemtic means gives $$(1+x)(1+y)(1+z)\le \left (1+{x+y+z\over 3}\right )^3$$ Next by the Cauchy-Schwarz inequality we get $$x+y+z\le \sqrt{3}\sqrt{x^2+y^2+z^2}\le \sqrt{3}$$ Thus $$(1+x)(1+y)(1+z)\le \left (1+{1\over\sqrt{3}}\right )^3\quad (*) $$ By analyzing the inequalities on the way the equality in $(*)$ is attained for $x=y=z={1\over \sqrt{3}}.$

Remark If it is compulsory to apply the multiplier method, we can equivalently maximize the function $$\ln[(1+x)(1+y)(1+z)]\\ =\ln(1+x)+\ln(1+y)+\ln(1+z)$$ In this way the equations take the form $${1\over 1+x}=2\lambda x,\ {1\over 1+y}=2\lambda y,\ {1\over 1+z}=2\lambda z $$ which should be easy for solving.

Answer 4

On the request of @TheoBendit, here's how I got my solutions:

Set up the Lagrange function: $\Lambda(x,y,z,\lambda) = f-\lambda g$ and set it's gradient to $0$ i.e $$\nabla \Lambda = \begin{bmatrix} (y+1) (z+1)-2 \lambda x \\ (x+1) (z+1)-2 \lambda y \\ (x+1) (y+1)-2 \lambda z \\ -x^2-y^2-z^2+1 \\ \end{bmatrix} =\textbf{0}$$ First, make $y$ the subject in equation $2$ to get $y=\frac{(x+1) (z+1)}{2 \lambda }$ and substitute this in equations $1$ and $3$ to get (with a bit of substitution): $$\frac{\overbrace{(2 \lambda +z+1)}^{B_1} \overbrace{(x z+x-2 \lambda x+z+1)}^{B_2}}{2 \lambda }=0\\ \frac{\overbrace{(2 \lambda +x+1)}^{B_3} \overbrace{(x z+x-2 \lambda z+z+1)}^{B_4}}{2 \lambda }=0$$ If $B_1=0$ and $B_3=0$, $x=z=-1-2\lambda$ so $y=2\lambda$. If $B_1=0$ and $B_4=0$, $x=y=-1-2\lambda$ and $z=2\lambda$. If $B_2=0$ and $B_3=0$, $y=z=-1-2\lambda$ and $x=2\lambda$.So our solutions are some permutation of $(-1-2\lambda,2\lambda,-1-2\lambda)$. Now plugging this information in equation 4 gives us: $$2(-1-2\lambda)^2+4\lambda^2=1\implies \lambda = -\frac{1}{2},-\frac{1}{6}$$ If $\lambda = -\frac{1}{2}$, out solutions turn out to be: $(x,y,z,\lambda)=(0,0,-1,-\frac{1}{2}),(0,-1,0,-\frac{1}{2}),(-1,0,0,-\frac{1}{2})$. If instead $\lambda=-\frac{1}{6}$, our solutions turn out to be $(x,y,z,\lambda)=(-\frac{2}{3},-\frac{2}{3},-\frac{1}{3},-\frac{1}{6}),(-\frac{2}{3},-\frac{1}{3},-\frac{2}{3},-\frac{1}{6}),(-\frac{1}{3},-\frac{2}{3},-\frac{2}{3},-\frac{1}{6})$.

The last combination is if $B_2=0$ and $B_4=0$. $$x z+x-2 \lambda x+z+1=0\implies x=\frac{-z-1}{-2 \lambda +z+1}$$ Subbing this in $B_4$ we get: $$\frac{\left(2 \lambda +\frac{-z-1}{-2 \lambda +z+1}+1\right) \left(\frac{z (-z-1)}{-2 \lambda +z+1}+\frac{-z-1}{-2 \lambda +z+1}-2 \lambda z+z+1\right)}{\lambda }=0 \\ \implies-\frac{4 \lambda (z-2 \lambda ) \left((z+1)^2-2 \lambda z\right)}{(-2 \lambda +z+1)^2}=0\\ \implies z=\lambda-1\pm\sqrt{\lambda ^2-2 \lambda }$$ This gives us $x=y=z=\lambda-1\pm\sqrt{\lambda ^2-2 \lambda}$. Subbing these in equation $4$ gives us $$3(\lambda-1\pm\sqrt{\lambda ^2-2 \lambda })^2=1 \implies \lambda = 1\mp\frac{2}{\sqrt{3}}$$ So our solutions become $(x,y,z,\lambda)=\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},1-\frac{2}{\sqrt{3}}\right)$ and $(x,y,z,\lambda)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},1+\frac{2}{\sqrt{3}}\right)$