Answer to Question #293781 in Microeconomics for Agyei Prince

Question

Find the turning point and dertemine whether they are Maxima or minima.

i.y=x2-6x+9

ii.y=x2+5xz+z2

iii.y=x2+z2+v2

Answer 1

i.

y'=(x2-6x+9'=2x-6

y''=(2x-6)''=2

2x-6=0

x=3

The second derivative does not contain x , so insertion gives 2. 2 is larger than 0. So there is a mimimum at 3.

$y(3)=3^2-6\times3+9=9-18+9=0$

Minimum turning point (3;0).

ii. y(x):

$x^2+5xz=y-z^2$

$x^2+5xz+(\frac{5z}{2})^2=(\frac{5z}{2})^2+y-z^2$

solution set

$x1=\frac{-5z}{2}+(y-z^2+\frac{25z^2}{4})^{0.5}$

$x2=\frac{-5z}{2}-(y-z^2+\frac{25z^2}{4})^{0.5}$

  y(z)

$z^2+5xz=-x^2+y$

$z^2+5xz+(\frac{5x}{2})^2=(\frac{5x}{2})^2+y-x^2$

$x1=\frac{-5x}{2}+(y-x^2+\frac{25x^2}{4})^{0.5}$

$x2=\frac{-5x}{2}-(y-x^2+\frac{25x^2}{4})^{0.5}$

iii.

y(x)

$z^2+v^2=y-x^2$

$z^2=y-x^2-v^2$

$z=+-(y-x^2-v^2)^{0.5}$

y(z)

$z^2+x^2=y-v^2$

$x^2=y-z^2-v^2$

$x=+-(y-z^2-v^2)^{0.5}$

y(v)

$z^2+v^2=y-x^2$

$v^2=y-z^2-x^2$

$v=+-(y-z^2-x^2)^{0.5}$