3 variables multiplication with 1 constraint lagrange multiplier

Question

The question states that : Suppose we want to optimise the function $f(x,y,z) = ax^2 + by^2 + cz^2$, where $a$, $b$, $c$ are positive constants, subject to the constraints $x^3 + y^3 + z^3 = 1$ and $x,y,z > 0$. Show that this problem has only one solution and the optimal value of $f(x,y,z)$ is equal to $(a^3 + b^3 + c^3)^{\frac 13}$.

My teacher just teaches us how to solve this type of problem with degree 1, but not of degree 2 and 3, so I have no idea how to solve this question. Please help me, thank you.

Answer 1

Hint: Solve the system $$2ax+3\lambda x^2=0$$ $$2by+3\lambda y^2=0$$ $$2cz+3\lambda z^2=0$$ $$x^3+y^3+z^3=1$$

Answer 2

Consider the Lagrange function with Lagrange multiplier $ \lambda: $ $$ \mathcal{L}(x,y,z,\lambda)=f(x,y,z)-\lambda(x^3+y^3+z^3-1) $$ satisfying the following equations: \begin{cases} \dfrac{\partial \mathcal{L}}{\partial x}=2ax-3\lambda x^2&=0\\ \dfrac{\partial\mathcal L}{\partial y}=2by-3\lambda y^2&=0\\ \dfrac{\partial\mathcal L}{\partial z}=2cz-3\lambda z^2&=0\\ \dfrac{\partial\mathcal L}{\partial \lambda}=-(x^3+y^3+z^3-1)&=0 \end{cases} Since $ a,b,c,x,y,z>0 $, we have $ \lambda\ne 0 $ and \begin{cases} x=\dfrac{2a}{3\lambda}\\ y=\dfrac{2b}{3\lambda}\\ z=\dfrac{2c}{3\lambda}\\ x^3+y^3+z^3=1 \end{cases} So you have \begin{align} 27\lambda^3=8(a^3+b^3+c^3) \end{align} i.e., $$ 3\lambda=2\sqrt[3]{a^3+b^3+c^3} $$ and \begin{cases} x=a(a^3+b^3+c^3)^{-\frac{1}{3}}\\ y=b(a^3+b^3+c^3)^{-\frac{1}{3}}\\ z=c(a^3+b^3+c^3)^{-\frac{1}{3}} \end{cases}

So the minimum value of $ f(x,y,z) $ constrained to the conditions is $$ f(a(a^3+b^3+c^3)^{-\frac{1}{3}},b(a^3+b^3+c^3)^{-\frac{1}{3}},c(a^3+b^3+c^3)^{-\frac{1}{3}}) $$ which is $$ (a^3+b^3+c^3)^{\frac{1}{3}}. $$

Answer 3

Calling

$$ L(x,y,z,\lambda) = ax^2+b y^2+z z^2-\lambda(x^3+y^3+z^3-1) $$

the stationary conditions are

$$ \begin{cases} 2ax-3\lambda x^2= 0\\ 2b y - 3\lambda y^2 = 0\\ 2c z - 3\lambda z^2 = 0\\ x^3+y^3+z^3= 1 \end{cases} $$

So we have

$$ \begin{cases} x = \frac{2a}{3\lambda}\\ y = \frac{2b}{3\lambda}\\ z = \frac{2c}{3\lambda}\\ \left(\frac{2a}{3\lambda}\right)^3+\left(\frac{2b}{3\lambda}\right)^3+\left(\frac{2c}{3\lambda}\right)^3=1 \end{cases} $$

or

$$ \sqrt[3]{a^3+b^3+c^3} = \frac{3}{2}\lambda $$

substituting back we have

$$ \frac{x^*}{a} = \frac{y^*}{b}=\frac{z^*}{c} = \frac{1}{\sqrt[3]{a^3+b^3+c^3}} $$

and thus

$$ f(x^*,y^*,z^*) = \frac{(a^3+b^3+c^3)}{\left(a^3+b^3+c^3\right)^{\frac{2}{3}}} = \left(a^3+b^3+c^3\right)^{\frac{1}{3}} $$