A number less eleven algebraic

Mathematics

1 answer:

bonufazy [111]3 years ago

5 0

Let the variable n represent the number

$\sf \longrightarrow \: n$

Less eleven means subtract eleven from the variable

$\sf \longrightarrow \: n - 11$

➪Thus, The equation is (n-11)

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Which trail is the closest in length to 10.5 kilometers?

Yuliya22 [10]

First of all you calculate the difference between the numbers:
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You then select the difference with the smallest number (ignoring the positive and negative symbols). That happens to be the last one.

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3 years ago

Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1

Varvara68 [4.7K]

I assume there are some plus signs that aren't rendering for some reason, so that the plane should be $x+y+z=1$ .

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The Lagrangian is

$L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)$

Take your partial derivatives and set them equal to 0:

$\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}$

Adding the first three equations together yields

$2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43$

and plugging this into the first three equations, you find a critical point at $(x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)$ .

The squared distance is then $d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43$ , which means the shortest distance must be $\sqrt{\dfrac43}=\dfrac2{\sqrt3}$ .