# Basic maths-calculus-Q17

+1 vote

If $x^{2} - 5x + 4 <= 0$ and $y^{2} -6y+5 <= 0$ what are the maximum and minimum values of $\frac{y+7x}{y+4x}$

A) 29/17 , 12/9

B) 23/13, 11/8

C) 25/11, 14/11

D) None of these

reshown May 24

$x^{2} - 5x + 4 <= 0$

So, $(x-1)(x-4) \leq 0$

So, $1 \leq x \leq 4$

and $y^{2} -6y+5 <= 0$

So, $(y-1)(y-5) \leq 0$

So, $1 \leq y \leq 5$

Given function :

$Z = \frac{y+7x}{y+4x}$

Now, we can check value of the function Z at (1,1), (1,5) , (4,1), (4,5)

$Z(1,1) = 1.6; \\ Z(1,5) = 12/9 = 1.33 ; \\ Z(4,1) = 29/17 = 1.7058; \\ Z(4,5) = 1.5714$

So, Z is maximum at (4,1) and maximum value is 29/17. Similarly, Z is minimum at (1,5) and minimum value is 12/9.

The function Z has No local maxima or local minima in the interval $1 \leq x \leq 4$ ; $1 \leq y \leq 5$  except at the corner points of the interval.

To see this we can apply the method "Max/min for functions of two variables" which is a tedius task.

First we find partial derivatives w.r.t  x and y.

∂f /∂x = $f_x$

$f_x = \frac{3y}{(y+4x)^2}$

∂f /∂y =    $f_y$

$f_y = \frac{-3x}{(y+4x)^2}$

A point (a, b) which is a maximum, minimum or saddle point is called a stationary point.

To find the stationary points of Z = f(x, y), work out  ∂f /∂x and ∂f /∂y and set both to zero. This gives you two equations for two unknowns x and y. Solve these equations for x and y

So, to find stationary points, we put $f_y , f_x$  both equal to zero and find stationary points.

$f_y =0$  means x = 0 ; $f_x =0$ means y = 0

So, only stationary point possible is  $(0,0)$  But at $(0,0)$ the function Z is Not defined, so, we can say that there is No stationary point for this function.

http://personal.ee.surrey.ac.uk/Personal/S.Zelik/teach/calculus/max_min_2var.pdf

answered Jun 3 by (112,760 points)
I got that x lies in [1,4] and y lies in [1,5], now in the first method how come you have checked on only those 4 points. There are infinite points in that range...
There is No stationary points for this function which means that this function will have maximum and minimum values at the corners of any interval because assume that there is a maximum value in the interval somewhere which is more than the values at corner points then there will be some stationary point(local maxima or minima or inflection point) which we have shown doesn't exist.