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Find the minimum of the function of two variable defined as

$$f(x_1,x_2)=(x_1-x_2)^2 +\left(2\sqrt2x_1-\sqrt{6(x_2)-(x_2)^2-7}\right)^2.$$

At first I tried to figure out a integral in this but failed. Then I tried using graphs, but things became complex. Any hints how to start the problem? Can I use partial differentiation?

user354545
  • 577

3 Answers3

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From $f'_{x_1}=0,\;f'_{x_2}=0$

we get

$ \left\{ \begin{array}{l} 2 x_1-2 x_2-\dfrac{2 \sqrt{2} \sqrt{-x_2^2+6 x_2-7}}{\sqrt{x_1}}+8=0 \\ -2 x_1+\dfrac{4 \sqrt{2 x_1} (x_2-3)}{\sqrt{-x_2^2+6 x_2-7}}+6=0 \\ \end{array} \right. $

Which holds for $x_1=-1;\;x_2=1$

and $f(-1,1)=2$ is the minimum because second derivative $f''_{xx}=-\frac{\sqrt{2} (6-2 x_2)}{\sqrt{x_1} \sqrt{-x_2^2+6 x_2-7}}-2$ and Hessian

$H(x_1,x_2)=-\frac{4 \left(4 \sqrt{2} x_1^{3/2}-x_1 \left(-x_2^2+6 x_2-7\right)^{3/2}-2 \sqrt{2} \sqrt{x_1} \left(x_2^3-9 x_2^2+25 x_2-21\right)+2 \left(-x_2^2+6 x_2-7\right)^{3/2}\right)}{x_1 \left(-x_2^2+6 x_2-7\right)^{3/2}}$

are positive at $(-1,1)$ so for the second derivative test $(-1,1)$ is a minimum.

Hope this helps

Raffaele
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Yes you will have to use Partial Differentiation to find and classify the maximum and minimum points on the surface.

JakeS4
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  • The process is too lengthy, isnt there any better way out. Some of my batchmates were saying they did this by graphical method – user354545 Nov 26 '17 at 17:14
  • @user354545 you could graph it using an online application, but to graph it by hand you would need to complete the same process to find the maximums and minimums. – JakeS4 Nov 26 '17 at 17:20
  • By partial differentiation? – user354545 Nov 26 '17 at 17:21
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show that $$f(x_1,x_2)\geq 11-6\sqrt{2}$$ and the equal sign holds if $$x_1=0,x_2=\frac{1}{6}(18-6\sqrt{2})$$

Dr. Sonnhard Graubner
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