Question

What is the shortest distance from the surface xy+12x+z^2=144 to the origin?

distance = ?

Answer 1

Let P(x,y,z) be the point on the surface $xy + 12x + z^{2} = 144$ . Then, distance between these two points is $d = \sqrt{x^2+y^2+z^2}$. We have to find the point (x, y, z) which is at a minimum distance from origin. We now use Lagrange's multipliers to find the shortest distance.

Let $f(x,y,z) = d^{2} = x^{2} + y^{2} + z^{2}$ and $g(x,y,z) = xy+12x+z^{2} = 144$.

Let

$F(x,y,z) = f(x,y,z) + \lambda g(x,y,z)\\ = x^{2} + y^{2}+z^{2}+\lambda(xy+12x+z^{2}-144)$

To find the point of shortest distance we have to solve,

$\dfrac{\partial F}{\partial x} = 0; \dfrac{\partial F}{\partial y} = 0; \dfrac{\partial F}{\partial z} = 0~ and ~\dfrac{\partial F}{\partial \lambda} = 0.$

$2x+\lambda(y+12) = 0\\ 2y+\lambda x=0\\ 2z+2\lambda z=0\\ xy+12x+z^2=144.$

From the third equation, we get $z=0~~\text{or}~~ \lambda =-1$. When $\lambda =-1$, first and second equation becomes,

$2x-y=12\\ -x+2y=0.$

Solving we get, $x= 8, y=4.$ From the constraint equation we get, $z = \pm 4.$

The critical points are $(8,4,-4), (8,4,4).$

Now when z = 0, the constraint becomes, $xy+12x=144.$

$x(y+12)=144\\ x=\dfrac{144}{y+12}.$Using this in f(x,y), we get

$f(x,y) = (\dfrac{144}{y+12})^{2}+y^2.$

Now we find the critical points of this one variable problem. We solve,

$\dfrac{\partial f}{\partial y} = 0\\ -\dfrac{2 \cdot 144^2}{(y+12)^3}+2y =0\\ y(y+12)^3-144^2=0\\ y^4+36y^3+432y^2+1728y-20736=0$

Solving this using online calculator, we get

$y=-9.366\pm10.973i\\ y=-21.83\\ y=4.5633$

Thus the critical points are, $(8.694,4.5633,0)$ and $(-14.65,-21.83,0)$ . The value of f at all critical points are,

$f(8,4,-4)=96\\ f(8,4,4)=96\\ f(8.694,4.5633,0)=96.41\\ f(-14.65,-21.83,0)=691.17$

Therefore, the shortest distance is $d = \sqrt{96} = 4\sqrt{6}$ which is attained at $(8,4,\pm4)$.