Question

What is the shortest distance from the surface xy+12x+z2=144 to the origin?

distance=?

Answer 1

Let the square of the distance function be $f(x,y,z)=x^2+y^2+z^2$

Given, the constraint function $g(x,y,z)=xy+12x+z^2$

Apply Lagrange Multiplier as,

$$\nabla f(x,y,z)=\lambda g(x,y,z)$$

So, find the gradient both sides to obtain,

$$<2x,2y,2z>=\lambda\ <y+12,x,2z>$$

Equate to gradients of both sides as,

$$2x=\lambda (y+12), 2y=\lambda (x), 2z=\lambda (2z)$$

From equation $2z=\lambda (2z),\lambda=1$ , so plug $\lambda=1$ into $2y=\lambda (x)$ to obtain $x=2y$.

Next, plug $x=2y$ and $\lambda=1$ into equation $2x=\lambda (y+12)$ and solve for $y$ as,

$$2(2y)=(y+12)$$ $$4y=y+12$$

Subtract $y$ from both sides to isolate variable $y$ from constant as,

$$4y-y=y+12-y$$ $$3y=12$$

Divide both sides by $3$ to obtain,

$$\frac{3y}{3}=\frac{12}{3}$$ $$y=4$$

So, the value of $y$ is $y=4$ and $x=2y=2(4)=8$

Now, plug the values of $x$ and $y$ into constraint equation $xy+12x+z^2=144$ and solve for $z$ as,

$$(8)(4)+12(8)+z^2=144$$ $$32+48+z^2=144$$ $$80+z^2=144$$

Subtract $80$ from both sides to obtain,

$$80+z^2-80=144-80$$ $$z^2=64$$ $$z=±8$$

Now, plug $x=8,y=4$ and $z=±8$ into objective function $f(x,y,z)=x^2+y^2+z^2$ to obtain,

$$f(8,4,±8)=(8)^2+(4)^2+(±8)^2=64+16+64=144$$

Next, find the square root of objective function as,

$$d=\sqrt{144}=12$$

Therefore, the shortest distance from the surface $xy+12x+z^2=144$ to the origin is $[12]$.