Lagrangian:

where the function we want to minimize is actually

, but it's easy to see that

and

have critical points at the same vector

.
Derivatives of the Lagrangian set equal to zero:




Substituting the first three equations into the fourth gives


Solving for

, we get a single critical point at

, which in turn gives the least distance between the plane and (0, 2, 5) of

.
Answer:
(8,-2)
Step-by-step explanation:
9-1 and 1-3
We don't need to graph the line. All we have to do is use our slope formula.
Answer is provided in the image attached.
Last one..................................................................................................................................................
Answer:
Area=91 ft^2
Step-by-step explanation:
Area=length*width
Area= 14*6.5
Area=91 ft^2