4
$\begingroup$

Find maximum and minimum value of function $$f(x,y,z)=y+z$$ on the circle $$x^2+y^2+z^2 = 1,3x+y=3$$

We have that $$y=3-3x$$ So we would like find minimum and maximum value of function $$g(x,z)=3-3x+z$$ on $$x^2+ (3(1-x))^2 + z^2 = 1$$ And then I was using Lagrange multipliers. I received that $$z=\frac{9-10x}{3}$$ and next I substituted it to circle equal. But the numbers are awful. This task comes from an exam so I suppose that is the easier way to solve it.

$\endgroup$
  • $\begingroup$ f(x,y,z) = z + y??? Where's x? $\endgroup$ – Stefan4024 Aug 31 '13 at 13:14
  • $\begingroup$ @Stefan4024: why does there need to be an $x$? $\endgroup$ – robjohn Aug 31 '13 at 14:23
  • $\begingroup$ Because f(x,y,z) means we have 3 variables. If we have just z and y as variables, then we can write it as: f(y,z) = y + z $\endgroup$ – Stefan4024 Aug 31 '13 at 14:34
  • $\begingroup$ @Stefan4024: but it is a function on $\mathbb{R}^3$ that doesn't depend on $x$. $\endgroup$ – robjohn Aug 31 '13 at 16:36
2
$\begingroup$

Let's look at what happens when we perturb $(x,y,z)$ infinitesimally in the direction $(\delta x,\delta y,\delta z)$.

To stay on $x^2+y^2+z^2=1$, we need $x\,\delta x+y\,\delta y+z\,\delta z=0$ and to stay on $3x+y=3$, we need $3\delta x+\delta y=0$.

That is, we want to consider only $(\delta x,\delta y,\delta z)$ which are perpendicular to both $(x,y,z)$ and $(3,1,0)$.

We want to find the point at which $\delta f(x,y,z)=(0,1,1)\cdot(\delta x,\delta y,\delta z)=0$ for all $(\delta x,\delta y,\delta z)$ which are perpendicular to both $(x,y,z)$ and $(3,1,0)$. For that to be true, $(0,1,1)$ must be a linear combination of $(3,1,0)$ and $(x,y,z)$. This means that $(x,y,z)$ is also a linear combination of $(3,1,0)$ and $(0,1,1)$; that is, on the plane $x-3y+3z=0$.

Thus, we are looking for a point on the unit sphere that is on the planes $x-3y+3z=0$ and $3x+y=3$. The intersection of these two planes is the line $(0,3,3)+(-3,9,10)t$. So we need to solve $$ \begin{align} 1&=(0-3t)^2+(3+9t)^2+(3+10t)^2\\ &=18+114t+190t^2 \end{align} $$ that is $t=\dfrac{-57\pm\sqrt{19}}{190}$. Note that $$ \begin{align} f((0,3,3)+(-3,9,10)t) &=(0,1,1)\cdot((0,3,3)+(-3,9,10)t)\\ &=6+19t \end{align} $$ Plugging in the values of $t$ give us the extreme values: $\dfrac{3+\sqrt{19}}{10}$ and $\dfrac{3-\sqrt{19}}{10}$

$\endgroup$
1
$\begingroup$

Another way. Given that you have $g = 3- 3x + z$ to find extrema, with the constraint

$1 = x^2 + 9(1-x)^2 + z^2 = (10x^2 -18x + 9) + z^2 = 10(x-\frac{9}{10})^2 + z^2 + \frac{9}{10}$

Thus the constraint is $100(x-\frac{9}{10})^2 + 10z^2 = 1$

Now we can let $x - \frac{9}{10} = \frac{\sin(t)}{10}$ and $z = \frac{\cos(t)}{\sqrt{10}}$ so that the constraint is satisfied.

Then our objective becomes $g = 3 - 3(\frac{9}{10}+\frac{\sin(t)}{10})+ \frac{\cos(t)}{\sqrt{10}}$.

$$\implies 10g-3 = - 3 \sin t + \sqrt{10} \cos t$$

Finding the extrema of this RHS should be easy using trigonometry or calculus, extrema of $a \sin t + b \cos t$ is $\pm \sqrt{a^2 + b^2}$. So we have

$$-\sqrt{19} \le 10g - 3 \le \sqrt{19}$$

$\endgroup$
0
$\begingroup$

Using Lagrange/Lagrangian multiplier, we have

$\displaystyle f(x,y,z)=y+z+a\{x^2+y^2+z^2-1\}$

$\displaystyle=3-3x+z+a\{x^2+(3-3x)^2+z^2-1\}=10ax^2-x(18a+3)+8a+az^2+3+8a$

$f_x=20ax-(18a+3)$

$f_z=1+2az$

$f_{xx}=20a,f_{zz}=2a,f_{xz}=f_{zx}=0$

Can you take it from here?

$\endgroup$
0
$\begingroup$

'Tis a pity you didn't continue, because you were nearly there. (It should be said that we can't expect the numbers to be too pretty, since there is rather limited symmetry among the function to be extremized and the constraint functions.) There are a couple of things we can do to keep the computations from getting messier than necessary.

Macavity shows the "constraint ellipse" that arises from the substitution you made; it "looks" very elongated as this is the projection onto the $ \ xz-$ plane of the intersection curve between the unit sphere and the plane $ \ 3x + y \ = \ 3 \ $ (the projection onto the $ \ yz-$ plane appears more nearly circular). From there, he pursued the path of parameterizing the intersection circle and the modified constraint function.

To follow your line of extremizing $ \ g(x,z) \ = \ 3 \ - \ 3x \ + \ z \ $ on the constraint ellipse, the Lagrange equations are

$$ -3 \ = \ \lambda \ \cdot \ 200 \ \left( x \ - \ \frac{9}{10} \right) \ \ , \ \ 1 \ = \ \lambda \ \cdot \ 20z \ \ ; $$

rather than multiply anything out, we will write

$$ \frac{1}{\lambda} \ = \ 20z \ = \ -\frac{200}{3} \left( x \ - \ \frac{9}{10} \right) \ \ \Rightarrow \ \ z \ = \ -\frac{10}{3} \left( x \ - \ \frac{9}{10} \right) \ \ , $$

which is what you've found, except we'll leave it in this form. We now insert this into the constraint equation to obtain

$$ 100 \left(x \ - \ \frac{9}{10} \right)^2 + \ 10 \left[ \ -\frac{10}{3} \left( x \ - \ \frac{9}{10} \right) \ \right]^2 \ = \ 1 $$

$$ \Rightarrow \ \ \left( 100 \ + \ \frac{1000}{9} \right) \ \left(x \ - \ \frac{9}{10} \right)^2 \ = \ 1 \ \ \Rightarrow \ \ \left(x \ - \ \frac{9}{10} \right)^2 \ = \ \frac{9}{1900} $$

$$ \Rightarrow \ \ x \ = \ \frac{9}{10} \ \pm \ \frac{3}{10 \ \sqrt{19}} \ = \ \frac{9 \cdot 19 \ \pm \ 3 \sqrt{19}}{190} \ \ . $$

With the values of $ \ x \ $ for the extremal points in hand, we can combine our results for $ \ y \ $ and $ \ z \ $ in order to compute the extremal values of our function:

$$ f(x, \ y, \ z) \ = \ y \ + \ z \ = \ (3 \ - \ 3x) \ + \ ( 3 \ - \ \frac{10}{3} x) \ = \ 6 \ - \ \frac{19}{3} x $$

$$ \rightarrow \ \ 6 \ - \ \frac{19}{3} \left( \frac{9 \cdot 19 \ \pm \ 3 \sqrt{19}}{190} \right) \ = \ 6 \ - \ \left( \frac{3 \cdot 19 \ \pm \ \sqrt{19}}{10} \right) \ = \ \frac{60 \ - \ 57 \ \pm \ \sqrt{19}}{10} \ = \ \frac{3 \ \pm \ \sqrt{19}}{10} \ \ [ \ \approx \ +0.736, \ -0.136 \ ] \ \ , $$

as also found in the other answers here. With a bit more (unasked-for) effort, we can calculate the extremal points as

$$ \left(\frac{171 \ - \ 3 \sqrt{19}}{190}, \ \frac{57 \ + \ 9 \sqrt{19}}{190}, \ \frac{\sqrt{19}}{19} \right) \ \approx \ (0.831, \ 0.506, \ 0.229) \quad \text{[maximum]} $$ and $$ \left(\frac{171 \ + \ 3 \sqrt{19}}{190}, \ \frac{57 \ - \ 9 \sqrt{19}}{190}, \ -\frac{\sqrt{19}}{19} \right) \ \approx \ (0.969, \ 0.094, -0.229) \quad \text{[minimum]} \ \ . $$

A graph with views of the geometrical situation is presented below.

enter image description here

$ \ $

Another method we can apply is the two-constraint form of Lagrange multipliers, in which we can define our constraint functions $ \ g(x, \ y, \ z) \ = \ x^2 \ + \ y^2 \ + \ z^2 \ - \ 1 \ $ and $ \ h(x, \ y, \ z) \ = \ 3x \ + \ y \ - \ 3 \ $ . We then introduce two multipliers in the equation $ \ \nabla f \ = \ \lambda \nabla g \ + \ \mu \nabla h \ $ to produce

$$ 0 \ = \ \lambda \ \cdot \ 2x \ + \ \mu \ \cdot \ 3 \ \ \ , \ \ \ 1 \ = \ \lambda \ \cdot \ 2y \ + \ \mu \ \cdot \ 1 \ \ \ , \ \ \ 1 \ = \ \lambda \ \cdot \ 2z \ + \ \mu \ \cdot \ 0 $$

$$ \Rightarrow \ \ \lambda \ = \frac{1}{2z} \ \ \Rightarrow \ \ \mu \ = \ 1 \ - \ \left( \frac{1}{2z} \right) \cdot 2y \ = \ -\frac{1}{3} \left( \frac{1}{2z} \right) \cdot 2x $$

$$ \Rightarrow \ \ \frac{y}{z} \ - \ \frac{x}{3z} \ = \ 1 \ \ \Rightarrow \ \ 3y \ - \ x \ = \ 3z \ \ , \ \ z \ \ne \ 0 \ \ . $$

Combining this with the constraint plane equation gives us $ \ z \ = \ y \ - \ \frac{1}{3} x \ = \ 3 \ - \ \frac{10}{3} x \ $ . Upon inserting this into the constraint sphere equation, we determine

$$ x^2 \ + \ (3 \ - \ 3x)^2 \ + \ \left( 3 \ - \ \frac{10}{3} x \right)^2 \ = \ 1 \ \ \Rightarrow \ \ \frac{190}{9} x^2 \ - \ 38 \ x \ + \ 17 \ = \ 0 \ \ , $$

for which the solutions are $ \ x \ = \ \frac{9 \cdot 19 \ \pm \ 3 \sqrt{19}}{190} \ $ , leading to the results found above.

[What I find interesting about this technique is that we can be fairly ignorant of the geometrical configuration and bypass much of the analytic geometry (for instance, we don't need to know where anything intersects). This is because the method will locate any points where there is a mutual tangency between the level surfaces of $ \ f(x, \ y, \ z) \ $ and the constraint surfaces, if such points exist.]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.