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.


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 $y=\frac{12}{5}$.

  • $\begingroup$ 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



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

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

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 $m:$



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