MHB Emwhy's question at Yahoo Answers regarding finding extrema in a given region

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Extrema
AI Thread Summary
The discussion centers on finding the extreme values of the function f(x,y) = 6x^2 + 3y^2 - 36x - 1 within the region defined by x^2 + y^2 ≤ 36, using methods such as Lagrange multipliers and critical point analysis. The critical point found inside the region is (3,0), which is identified as a relative minimum. The boundary is examined using parametric equations, leading to critical points at (6,0) and (-6,0). The analysis concludes that the absolute maximum value is 431 at (-6,0) and the absolute minimum value is -55 at (3,0). This comprehensive approach effectively identifies the extrema within the specified region.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

Find the extreme values of f, on a given region? Using LaGrange multipliers?


Find the extreme values of f(x,y) = 6x^2+3y^2-36x-1, on the region described by x^2+y^2 <=36.

I'm pretty sure you have to use LaGrange multipliers. I ended up with lamba=0, x=3, y= +/-sqrt(27) which doesn't seem right. Additional Details

Actually I just got the absolute minimum value by taking partial derivatives of f with respect to x and y. How do I get the absolute max?

I have posted a link there to this thread so the OP can see my work.
 
Mathematics news on Phys.org
Re: emwhy's question at Yahoo! Questions regarding finding extrema in a given region

Hello emwhy,

First, let's find the critical values inside the given region. We do this by equating the partials to zero, and solving the resulting system:

We are given:

$$f(x,y)=6x^2+3y^2-36x-1$$

Hence:

$$f_x(x,y)=12x-36=12(x-3)=0$$

$$f_y(x,y)=6y=0$$

This gives us the critical point $(x,y)=(3,0)$. Since its distance from the origin is less than $6$, we know it is within the circular region.

Next, we may employ the second partials test to determine the nature of the critical point.

Let:

$$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left(f_{xy}(x,y) \right)^2$$

Thus:

$$D(x,y)=12\cdot6-\left(0 \right)^2=72>0$$

Since $$f_{xx}(3,0)=12>0$$ we conclude that the critical point $(3,0)$ is a relative minimum.

Now we want to check the boundary. In order to examine $f$ on the boundary of the region, we may represent the circle $x^2+y^2=6^2$ by means of the parametric equations $x=6\cos(t),\,y=6\sin(t),\,0\le t\le 2\pi$. Thus, on the boundary we can write $f$ as a function of a single variable $t$:

$$f(\cos(t),\sin(t))=f(t)=6\left(6\cos(t) \right)^2+3\left(6\sin(t) \right)^2-36\left(6\cos(t) \right)-1=108\cos^2(t)-216\cos(t)+107$$

Differentiating with respect to $t$ and equating to zero, we find:

$$f'(t)=-216\cos(t)\sin(t)+216\sin(t)=216\sin(t)\left(1-\cos(t) \right)=0$$

This gives us the critical values for $t$:

$$t=0,\,\pi,\,2\pi$$

Since the first and third critical values are boundary values for $t$, and because the parametrization of $x$ and $y$ have the same value for these two values, we need only consider:

$$t=0,\,\pi$$

This gives us the two critical points:

$$\left(x(0),y(0) \right)=(6,0)$$

$$\left(x(\pi),y(\pi) \right)=(-6,0)$$

Now, we may use the second derivative test to determine the nature of these extrema. Let's write the first derivative as:

$$f'(t)=-108\sin(2t)+216\sin(t)$$

Differentiating, we find:

$$f''(t)=-216\cos(2t)+216\cos(t)$$

Checking the critical values:

$$f''(0)=0$$

The second derivative test is inconclusive for this point, so using the first derivative test, we find:

$$f'\left(\frac{\pi}{2} \right)=216>0$$

$$f'\left(\frac{3\pi}{2} \right)=-216<0$$

Hence the point $(6,0)$ is a relative minimum.

$$f''(\pi)=-432<0$$

Hence the point $(-6,0)$ is a relative maximum.

So, we now know:

$$f_{\max}=f(-6,0)=6(-6)^2+3(0)^2-36(-6)-1=431$$

$$f_{\min}=f(3,0)=6(3)^2+3(0)^2-36(3)-1=-55$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top