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Home Partial Differentiation Maxima and Minima Examples Example 5: Closest Distance
 
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Example 5: Closest Distance

Find the point on the plane

4x - 6y + 2z = 7

which is nearest to the origin.

Step 1

The distance from the origin to (x, y, z) is 11_partial_differentiation-470.gif. It is easier work with the square of the distance, which has a minimum at the same poi that the distance does. So we wish to find the minimum of

w = x2 + y2 + z2

given that

4x - 6y + 2z = 7.

We eliminate z using the plane equation.

z = ½(7 - 4x + 6y), w = x2 + y2 + ¼(7 - 4x + 6y)2.

The domain is the whole (x, y) plane.

Step 2

11_partial_differentiation-471.gif = 2x + 2·¼(-4)(7 - 4x + 6)-) = - 14 + 10x - 12y.

11_partial_differentiation-472.gif = 2y + 2 · ¼ · 6(7 - 4x + 6y) = 21 - 12x + 20y.

Step 3

-14 + 10x - 12y = 0, 21 - 12x + 20y = 0.

Solving for x and y we get one critical point

x = ½, y = -¾

CONCLUSION

We know from geometry that there is a point on the plane which is closest to the origin (the point where a perpendicular line from the origin meets the plane). Therefore w has a minimum and it must be at the critical point

x = ½, y = -¾

The value of z at this point is

z = ½(7 - 4x + 6y) = ¼

The answer is

(½, -¾ , ¼).

The plane is shown in Figure 11.7.10.

11_partial_differentiation-473.gif

Figure 11.7.10
Home Partial Differentiation Maxima and Minima Examples Example 5: Closest Distance

Last Update: 2006-11-15