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jeka57 [31]
3 years ago
12

Please help and show work if possible. thank you!!

Mathematics
1 answer:
Elena-2011 [213]3 years ago
3 0
10. Horizontal shift of 50, vertical shift of -20, horizontal shift of -50. Think of it on a plane, with right in the positive x-axis and up in the positive y-axis. The cans go right 50ft, then down 20ft, then left 50ft. In terms of the horizontal and vertical, they go 50ft in the positive horizontal axis, then 20ft in the negative vertical axis, then 50ft in the negative horizontal axis. Therefore, the cans have a horizontal shift of 50, then a vertical shift of -20, then a horizontal shift of -50.

11. Since the partition and the wall are parallel, the triangles are similar. This means that the ratio between the sides are the same for the small triangle and the big triangle. The small triangle (made by the partition) is 3m wide and 2m tall. Since the big triangle (made by the wall) is 4m tall, the sides of the big triangle are twice the size of the small triangle. Therefore, the big triangle is 6m wide. We cannot forget to subtract the 3m from the small triangle, since we only want to know how far the partition is from the wall, not how far the point is from the wall.
The wall is 3m away from the partition.
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The Lagrangian for this function and the given constraints is

L(x,y,z,\lambda_1,\lambda_2)=x^2+y^2+z^2+\lambda_1(x+2y+z-4)+\lambda_2(x-y-8)

which has partial derivatives (set equal to 0) satisfying

\begin{cases}L_x=2x+\lambda_1+\lambda_2=0\\L_y=2y+2\lambda_1-\lambda_2=0\\L_z=2z+\lambda_1=0\\L_{\lambda_1}=x+2y+z-4=0\\L_{\lambda_2}=x-y-8=0\end{cases}

This is a fairly standard linear system. Solving yields Lagrange multipliers of \lambda_1=-\dfrac{32}{11} and \lambda_2=-\dfrac{104}{11}, and at the same time we find only one critical point at (x,y,z)=\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right).

Check the Hessian for f(x,y,z), given by

\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}

\mathbf H is positive definite, since \mathbf v^\top\mathbf{Hv}>0 for any vector \mathbf v=\begin{bmatrix}x&y&z\end{bmatrix}^\top, which means f(x,y,z)=x^2+y^2+z^2 attains a minimum value of \dfrac{480}{11} at \left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right). There is no maximum over the given constraints.
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This graph will start at (.5, 0) as the vertex.

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You can also plot the other side of the graph by going up two and to the right one.

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