# Maximum value of a function with 2 variables

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:

$$\hspace{4cm}$$

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

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

$$y/x=m.$$

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

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

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

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 $$m:$$

$$m=3/4.$$