0
$\begingroup$

Can someone help me finding maximum value of a ratio in quadratic function in 2 variables using proper mathematical methods.?

Question is as below.

If x and y are real numbers such that $x^2 -10x+y^2 +16=0$, determine the maximum value of the ratio $y/x$

I know there is Ramban method to solve this. Taking $y/x=k --> y=kx$ and forming equation in x , then applying $^2 - 4ac >=0$ for max min value of k.

Is there any way to using differentiation ?

Sorry in advance if this is a repeat. I am new to platform.

$\endgroup$
2
$\begingroup$

Note that the equation is a circle with center $O(5,0)$ and radius $3$: $$x^2 -10x+y^2 +16=0 \iff (x-5)^2+y^2=9$$ The objective function is $\frac yx=k \iff y=kx$, whose contour lines will pass through the origin. So you need to find the slope of the tangent to the circle. See the graph:

$\hspace{4cm}$enter image description here

Hence, the slope is $k=\frac 34$, which is the maximum value of $\frac yx$ at $x=\frac{16}{5}$ and $y=\frac{12}{5}$.

$\endgroup$
  • $\begingroup$ Amazing. This is the kind of solution I was looking for that I can apply in aptitude exam within limited time. Thanks $\endgroup$ – Vishu Sahni Dec 13 '18 at 21:40
1
$\begingroup$

$y/x=m.$

$x^2-10x+(mx)^2+16=0.$

$(1+m^2)x^2 -10x+16=0.$

$\small{(1+m^2)\left (x^2-\dfrac{10}{1+m^2}x \right) +16=0.}$

Completing the square:

$\small{(1+m^2)\left (x-\dfrac{5}{1+m^2}\right)^2 -\dfrac{25}{1+m^2}+16=0.}$

$\small{(1+m^2)^2 \left (x-\dfrac{5}{1+m^2}\right )^2 =-16(1+m^2)+25 \ge 0.}$

Hence :

$25/16 \ge 1+m^2$.

$9/16 \ge m^2$.

Maximal $m:$

$m=3/4.$

$\endgroup$

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.