Use lagrange multipliers to find the point on the plane x − 2y + 3z = 6 that is closest to the point (0, 2, 5).

The point on the plane x − 2y + 3z = 6 that is closest to the point (0, 2, 5) is.

Let (x,y,z) be the point on the plane x − 2y + 3z = 6 that is closest to the point (0, 2, 5).Then

The gradient of f and g are

Δ=

Δ

Setting Δf=kΔg

k(1,-2,3)

2x=k,2(y-3)=-2k,2(z-5)=3k

Solving we get the point

Therefore the point on the plane x − 2y + 3z = 6 that is closest to the point (0, 2, 5) is.

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infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions

1e+27

step-by-step explanation: