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3 years ago

Use lagrange multipliers to find the points on the given surface that are closest to the origin. y2 = 64 + xz

1 answer:

8 0

The distance between an arbitrary point on the surface and the origin is

$d(x,y,z)=\sqrt{x^2+y^2+z^2}$

Recall that for differentiable functions $g(x)$ and $h(x)$ , the composition $g(h(x))$ attains extrema at the same points that $h(x)$ does, so we can consider an augmented distance function

$D(x,y,z)=x^2+y^2+z^2$

The Lagrangian would then be

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(y^2-64-xz)$

We have partial derivatives

$\begin{cases}L_x=2x-\lambda z\\L_y=2y+2y\lambda\\L_z=2z-\lambda x\\L_\lambda=y^2-64-xz\end{cases}$

Set each partial derivative to 0 and solve the system to find the critical points.

From the second equation it follows that either $y=0$ or $\lambda=-1$ . In the first case we arrive at a contradiction (I'll leave establishing that to you). If $\lambda=-1$ , then we have

$\begin{cases}2x+z=0\\2z+x=0\end{cases}\implies x=0,z=0$

This means $y^2=64\implies y=\pm8$

so that the points on the surface closest to the origin are $(0,\pm8,0)$ .

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Two lighthouses are located 75 miles from one another on a north-south line. If a boat is spotted S 40o E from the northern ligh

yuradex [85]

Answer:

The northern lighthouse is approximately $24.4\; \rm mi$ closer to the boat than the southern lighthouse.

Step-by-step explanation:

Refer to the diagram attached. Denote the northern lighthouse as $\rm N$ , the southern lighthouse as $\rm S$ , and the boat as $\rm B$ . These three points would form a triangle.

It is given that two of the angles of this triangle measure $40^{\circ}$ (northern lighthouse, $\angle {\rm N}$ ) and $21^{\circ}$ (southern lighthouse $\angle {\rm S}$ ), respectively. The three angles of any triangle add up to $180^{\circ}$ . Therefore, the third angle of this triangle would measure $180^{\circ} - (40^{\circ} + 21^{\circ}) = 119^{\circ}$ (boat $\angle {\rm B}$ .)

It is also given that the length between the two lighthouses (length of $\rm NS$ ) is $75\; \rm mi$ .

By the law of sine, the length of a side in a given triangle would be proportional to the angle opposite to that side. For example, in the triangle in this question, $\angle {\rm B}$ is opposite to side $\rm NS$ , whereas $\angle {\rm S}$ is opposite to side ${\rm NB}$ . Therefore:

$\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of NB}}{\sin(\angle {\rm S})} \end{aligned}$ .

Substitute in the known measurements:

$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of NB}}{\sin(21^{\circ})} \end{aligned}$ .

Rearrange and solve for the length of $\rm NB$ :

$\begin{aligned} & \text{length of NB} \\ =\; & (75\; \rm mi) \times \frac{\sin(21^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 30.73\; \rm mi\end{aligned}$ .

(Round to at least one more decimal places than the values in the choices.)

Likewise, with $\angle {\rm N}$ is opposite to side ${\rm SB}$ , the following would also hold:

$\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of SB}}{\sin(\angle {\rm N})} \end{aligned}$ .

$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of SB}}{\sin(40^{\circ})} \end{aligned}$ .

$\begin{aligned} & \text{length of SB} \\ =\; & (75\; \rm mi) \times \frac{\sin(40^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 55.12\; \rm mi\end{aligned}$ .

In other words, the distance between the northern lighthouse and the boat is approximately $30.73\; \rm mi$ , whereas the distance between the southern lighthouse and the boat is approximately $55.12\; \rm mi$ . Hence the conclusion.