Answer:
Step-by-step explanation:#1 = 39
#2=30
#3=black
Minimize
![(x-16)^2+(y-6)^2+z^2](https://tex.z-dn.net/?f=%28x-16%29%5E2%2B%28y-6%29%5E2%2Bz%5E2)
subject to
![z^2=x^2+y^2](https://tex.z-dn.net/?f=z%5E2%3Dx%5E2%2By%5E2)
. The Lagrangian would be
![L(x,y,z,\lambda)=(x-16)^2+(y-6)+z^2+\lambda(z^2-x^2-y^2)](https://tex.z-dn.net/?f=L%28x%2Cy%2Cz%2C%5Clambda%29%3D%28x-16%29%5E2%2B%28y-6%29%2Bz%5E2%2B%5Clambda%28z%5E2-x%5E2-y%5E2%29)
and has partial derivatives
![\begin{cases}L_x=2(x-16)-2\lambda x\\L_y=2(y-6)-2\lambda y\\L_z=2z+2\lambda z\\L_\lambda z^2-x^2-y^2\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7DL_x%3D2%28x-16%29-2%5Clambda%20x%5C%5CL_y%3D2%28y-6%29-2%5Clambda%20y%5C%5CL_z%3D2z%2B2%5Clambda%20z%5C%5CL_%5Clambda%20z%5E2-x%5E2-y%5E2%5Cend%7Bcases%7D)
Setting each partial derivative to 0, we have
![\begin{cases}2(x-16)-2\lambda x=0\implies (1-\lambda)x=16\\2(y-6)-2\lambda y=0\implies(1-\lambda)y=6\\2z+2\lambda z=0\implies(1+\lambda)z=0\\z^2-x^2-y^2=0\implies z^2=x^2+y^2\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D2%28x-16%29-2%5Clambda%20x%3D0%5Cimplies%20%281-%5Clambda%29x%3D16%5C%5C2%28y-6%29-2%5Clambda%20y%3D0%5Cimplies%281-%5Clambda%29y%3D6%5C%5C2z%2B2%5Clambda%20z%3D0%5Cimplies%281%2B%5Clambda%29z%3D0%5C%5Cz%5E2-x%5E2-y%5E2%3D0%5Cimplies%20z%5E2%3Dx%5E2%2By%5E2%5Cend%7Bcases%7D)
From the third equation, it follows that either
![\lambda=-1](https://tex.z-dn.net/?f=%5Clambda%3D-1)
or
![z=0](https://tex.z-dn.net/?f=z%3D0)
. In the second case, we arrive at a contradiction:
![z^2=x^2+y^2=0\implies x=y=0](https://tex.z-dn.net/?f=z%5E2%3Dx%5E2%2By%5E2%3D0%5Cimplies%20x%3Dy%3D0)
since both
![x^2](https://tex.z-dn.net/?f=x%5E2)
and
![y^2](https://tex.z-dn.net/?f=y%5E2)
must be non-negative, yet this would mean e.g.
![(1-\lambda)x=0=16](https://tex.z-dn.net/?f=%281-%5Clambda%29x%3D0%3D16)
. So it must be that
![\lambda=-1](https://tex.z-dn.net/?f=%5Clambda%3D-1)
.
The first and second equations then tell us that
![(1-\lambda)x=2x=16\implies x=8](https://tex.z-dn.net/?f=%281-%5Clambda%29x%3D2x%3D16%5Cimplies%20x%3D8)
![(1-\lambda)y=2y=6\implies y=3](https://tex.z-dn.net/?f=%281-%5Clambda%29y%3D2y%3D6%5Cimplies%20y%3D3)
from which we obtain
![z^2=8^2+3^2=73\implies z=\pm\sqrt{73}](https://tex.z-dn.net/?f=z%5E2%3D8%5E2%2B3%5E2%3D73%5Cimplies%20z%3D%5Cpm%5Csqrt%7B73%7D)
.
Thus the points on the cone closest to (16, 6, 0) are
![(8,3,\pm\sqrt{73})](https://tex.z-dn.net/?f=%288%2C3%2C%5Cpm%5Csqrt%7B73%7D%29)
.
Answer:
64
Step-by-step explanation:
Perfect squares are integers multiplied by themselves.
- 2 times 2 = 4
- 3 times 3 = 9
- 4 times 4 = 15
The closest perfect squares to 54 are 49 (7^2) and 64 (8^2).
49 is less than 54, so that's ruled out.
Therefore, the closest perfect square to 54 that is greater than it is 64.
It should be A if I did the math right which I think I did