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Find the maximum value of $\sqrt{x^2+y^2}$ where $$x^2+y^2=6x-8y+11$$ where $x,y$ are real numbers.

Now one approach is just to write the given expression as the equation of a circle, and the maximum value of $x^2+y^2$ would be the distance between origin and the farthest point on the circumference of the circle, which yields $11$ as the answer. But I was wondering if there's a way to do this problem using partial derivatives. I calculated the partial derivative of $x^2+y^2-6x+8y-11$ with respect to $y$ to be $2(y+4)$ and with respect to $x$ to be $2(x-3)$. How do I proceed?

Ellie_Wong
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2 Answers2

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You can use the method of the Lagrange multipliers. This method is based on the clever use of partial derivatives of a suitable target function. That is what the OP requested.

The function to be optimized is given by:

$$f(x,y, \lambda) = \sqrt{x^2+y^2} - \lambda(x^2 + y^2 -6x + 8y-11)$$

We now take partial derivatives of $f$ with respect to $x$, $y$ and $\lambda$ and set them equal to zero. This yields these $3$ equations:

$$x/\sqrt{x^2+y^2} -\lambda(2x-6)=0$$

$$y/\sqrt{x^2+y^2} -\lambda(2y+8)=0$$

$$x^2+y^2-6x+8y-11=0$$

Eliminating $\lambda$ from the first two equations yields $4x = -3y$. Substitute this relation into the third equation and you find that one solution is $x=-3/5$ and $y=4/5$. This corresponds to the minimum value of $\sqrt{x^2+y^2}$ which is $1$. The other solution is $x=33/5$ and $y=-44/5$. This corresponds to the maximum value of $\sqrt{x^2+y^2}$ which is $11$.

M. Wind
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Observe that

$$x^2+y^2=6x-8y+11\implies (x-3)^2+(y+4)^2=36$$

so that we have a circle with centre at $\;(3,-4)\;$ and radius $\;6\;$ . Can you take it from here?

DonAntonio
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  • I mentioned in my post that I have done this and I have got the correct answer. It's that I want to do it differently using partial derivatives specifically. – Ellie_Wong May 30 '23 at 13:07
  • Oh, I see. Well, then you try Lagrange multipliers with $$f(x,y,\lambda)=\sqrt{x^2+y^2}+\lambda(x^2+y^2-6x+8y-11)$$ Take partial first derivatives, equal them to zero and etc. – DonAntonio May 30 '23 at 13:11
  • I tried taking the partial derivative of the expression $x^2+y^2-6x+8y-11$ with respect to $x$ and $y$ and got $2(x-3), 2(y+4)$ respectively. I equalled them to $0$ which gives $(x,y)=(3,-4)$. But this doesn't get the correct answer. Can you tell me what's wrong in this approach? – Ellie_Wong May 30 '23 at 13:19