# Thread: Solving the equations arising from a Lagrange multipliers problem

1. ## Solving the equations arising from a Lagrange multipliers problem

Find the maximum value of 2
x+3y+7z, subject to the constraint x^4+y^4+z^4= 1.

First I formed my Lagrangian for this problem:

L(x,y,z) = 2x + 3y + 7z - λ(x^4 + y^4 + z^4 -1)

I then took my partial derivatives:

dL/dx = 2 - 4
λx^3

dL/dy = 3 - 4λy^3
dL/dz = 7 - 4λz^3

For each of these I made x,y and z the subject respectively, giving:

x^3 = 1/2λ ; y^3 = 3/4λ ; z^3 = 7/4λ

I now subbed these into my constraint. This gave a scary equation with the above λ expressions raised to the power of 4/3 being equal to 1. This is where I'm stuck, can anyone help me please?

I found a solution, but not in surd form, 2.38534589 satisfies the λ equation being equal to 1. But my answer must be in the form of a surd

2. Originally Posted by mitch_nufc
Find the maximum value of 2

x+3y+7z, subject to the constraint x^4+y^4+z^4= 1.

First I formed my Lagrangian for this problem:

L(x,y,z) = 2x + 3y + 7z - λ(x^4 + y^4 + z^4 -1)

I then took my partial derivatives:

dL/dx = 2 - 4λx^3
dL/dy = 3 - 4λy^3
dL/dz = 7 - 4λz^3

For each of these I made x,y and z the subject respectively, giving:

x^3 = 1/2λ ; y^3 = 3/4λ ; z^3 = 7/4λ

I now subbed these into my constraint. This gave a scary equation with the above λ expressions raised to the power of 4/3 being equal to 1. This is where I'm stuck, can anyone help me please?

I found a solution, but not in surd form, 2.38534589 satisfies the λ equation being equal to 1. But my answer must be in the form of a surd
i'm just wondering why you thought that this is an "abstract algebra" problem and definitely not a "calculus" problem!!??

3. Because I dont have a problem with the calculus, I have a problem solving the equation that comes from the calculus, which is an algebra problem.

4. Originally Posted by mitch_nufc
Find the maximum value of 2

x+3y+7z, subject to the constraint x^4+y^4+z^4= 1.

First I formed my Lagrangian for this problem:

L(x,y,z) = 2x + 3y + 7z - λ(x^4 + y^4 + z^4 -1)

I then took my partial derivatives:

dL/dx = 2 - 4λx^3
dL/dy = 3 - 4λy^3
dL/dz = 7 - 4λz^3

For each of these I made x,y and z the subject respectively, giving:

x^3 = 1/2λ ; y^3 = 3/4λ ; z^3 = 7/4λ

I now subbed these into my constraint. This gave a scary equation with the above λ expressions raised to the power of 4/3 being equal to 1. This is where I'm stuck, can anyone help me please?

I found a solution, but not in surd form, 2.38534589 satisfies the λ equation being equal to 1. But my answer must be in the form of a surd
Doing what you say you get:

$\left( \frac{1}{2} \lambda \right)^{4/3} + \left( \frac{3}{4} \lambda \right)^{4/3} + \left( \frac{7}{4} \lambda \right)^{4/3} = 1$

$\Rightarrow \lambda^{4/3} \left( \left(\frac{1}{2}\right)^{4/3} + \left( \frac{3}{4} \right)^{4/3} + \left( \frac{7}{4}\right)^{4/3}\right) = 1$

Now make $\lambda$ the subject. I don't see any simple surd form here but obviously there is a complicated surd form.

5. Originally Posted by mitch_nufc
Find the maximum value of 2
x+3y+7z, subject to the constraint x^4+y^4+z^4= 1.

First I formed my Lagrangian for this problem:

L(x,y,z) = 2x + 3y + 7z - λ(x^4 + y^4 + z^4 -1)

I then took my partial derivatives:

dL/dx = 2 - 4
λx^3

dL/dy = 3 - 4λy^3
dL/dz = 7 - 4λz^3

each of those is equal to 0, of course.

For each of these I made x,y and z the subject respectively, giving:

x^3 = 1/2λ ; y^3 = 3/4λ ; z^3 = 7/4λ

I now subbed these into my constraint. This gave a scary equation with the above λ expressions raised to the power of 4/3 being equal to 1. This is where I'm stuck, can anyone help me please?

I found a solution, but not in surd form, 2.38534589 satisfies the λ equation being equal to 1. But my answer must be in the form of a surd
It would have made more sense if you had made $\lambda$ the subject of each equation, the set them all equal: $\lambda= \frac{1}{2x^3}= \frac{3}{4y^3}= \frac{7}{4y^3}$. Or, you can eliminate $\lambda$ by dividing one equation by another. From $4\lambda x^3= 2$ and $4\lambda y^4= 3$, we get $\frac{4\lambda x^3}{4\lambda y^3}= \frac{x^3}{y^3}= \frac{2}{3}$ so that $x^3= \frac{2}{3}y^3$ and from $4\lambda y^3= 3$ and $4\lambda z^3= 7$ we get $\frac{4\lambda z^3}{4\lambda y^3}= \frac{z^3}{y^3}= \frac{7}{3}$ so $z^3= \frac{7}{3}y^3$. Now $x= \sqrt[3]{2/3}y$ so $x^4= (2/3)^{4/3}y^4$ and $z= \sqrt[3]{7/3}y$ so $z^4= (7/3)^{4/3}y^4$. You can put those into the constraint equation to get a single equation for y, and then get x and z.

6. Really need to start checking what you've posted before I post Matthew looks like what I've done thus far is right. I just thought it was a bit too complicated. It's far more complicated than the one on the examples sheet we were given, because the constraint there was only $x^2+y^2+z^2=1$ :\

7. Hello, mitch_nufc!

Find the maximum value of $2x+3y +7z$, subject to the constraint $x^4 + y^4 + z^4 \:=\: 1$

First I formed my Lagrangian for this problem:

$L(x,y,z,\lambda) \:= \:2x + 3y + 7z - \lambda(x^4 + y^4 + z^4 -1)$

I then took my partial derivatives:

. . $\begin{array}{ccccc}\dfrac{dL}{dx} &=& 2 - 4\lambda x^3 &=& 0 \\ \\[-3mm]
\dfrac{dL}{dy} &=& 3 - 4\lambda y^3 &=& 0 \\ \\[-3mm]
\dfrac{dL}{dz} &=& 7 - 4\lambda z^3 &=& 0 \end{array}$

I would solve for $\lambda\!:\quad \lambda \;=\;\frac{1}{2x^3} \;=\;\frac{3}{4y^3} \;=\;\frac{7}{4z^3}$

We have: . $14x^3 \:=\:4z^3 \quad\Rightarrow\quad x \:=\:\left(\tfrac{2}{7}z^3\right)^{\frac{1}{3}} \quad\Rightarrow\quad x^4 \:=\:\left(\tfrac{2}{7}\right)^{\frac{4}{3}}\!z^4$ .[1]

We have: . $28y^3 \:=\:12z^3 \quad\Rightarrow\quad y\:=\:\left(\tfrac{3}{7}z^3\right)^{\frac{1}{3}} \quad\Rightarrow\quad y^4 \:=\:\left(\tfrac{3}{7}\right)^{\frac{4}{3}}\!z^4$ .[2]

Substitute [1] and [2] into the constraint: . $\left(\tfrac{2}{7}\right)^{\frac{4}{3}}\!z^4 + \left(\tfrac{3}{7}\right)^{\frac{4}{3}}\!z^4 + z^4 \:=\:1$

Factor: . $\bigg[\left(\tfrac{2}{7}\right)^{\frac{4}{3}} + \left(\tfrac{3}{7}\right)^{\frac{4}{3}} + 1\bigg]\,z^4 \;=\;1 \quad\Rightarrow\quad\bigg[\frac{2^{\frac{4}{3}} + 3^{\frac{4}{3}} + 7^{\frac{4}{3}}}{7^{\frac{4}{3}}}\bigg]\,z^4 \;=\;1$

. . $z^4 \;=\;\frac{7^{\frac{4}{3}}}{2^{\frac{4}{3}} + 3^{\frac{4}{3}} + 7^{\frac{4}{3}}} \quad\Rightarrow\quad z \;=\;\frac{7^{\frac{1}{3}}} {\left(2^{\frac{4}{3}} + 3^{\frac{4}{3}} + 7^{\frac{4}{3}}\right)^{\frac{1}{4}}}$

And we can back-substitute to determine $x$ and $y.$