00:01
For this problem, we are asked to find the minimum distance from the point 1 ,23 to the plane 2x plus 3y plus z equals 12.
00:07
So we have the hint here.
00:09
To simplify the computations, we want to minimize the square of the distance.
00:14
So we'll have d squared equals f of xy, which in turn will be equal to x minus 1 squared plus y minus 2 squared plus i'll write this as z of xy minus 6.
00:30
Squared, where we can find z of xy by rearranging the equation of our plane to get z as a function of x and y.
00:38
Specifically, we get z of xy will be equal to 12 minus 2x minus 3y.
00:43
So our function then, f of xy, equals x minus 1 squared plus y minus 2 squared plus 12 minus 3, so 9, so plus 9 minus 2x minus 3 y all squared.
01:01
Now, let's see here.
01:05
What we want to do now is take the partial derivative of f with respect to x and y and set both to be equal to zero.
01:11
The partial derivative of f with respect to x after simplifying it down will be 10x plus 12y minus 38...