Find the point (x0, y0) on the line 12x+12y=2 that is closest to the origin.
I set ∇f to ∇g to get:
<2x,2y> = λ<12,12>
I then found λ = x/6 and λ = y/6, meaning that x = y
Once I plugged that back into the constraint, I found the answer to be:(1/12,1/12)
However, my online homework is saying my answer's incorrect. Did I do something wrong?
First reduce and change 12x+12y=2 to
6x + 6y = 1
Doing it another way:
slope of this line is -1
so the slope of the perpendicular is +1
and the line passing through the origin with slope 6 is y = x
sub into the given line 6x + 6y = 1
6x + 6x = 1
x = 1/12
y = 1/12
check is (1/12,1/12) on 12x+12y=2 ?
LS = 12(1/12) + 12(1/12)
= 1+1 = 2 = RS
My answer and your answer is correct, the book is wrong.
Based on your description of the steps you followed, it seems there might be a minor error in your calculations. Let's go through the solution together step by step to find the closest point on the line to the origin.
We start with the equation of the line:
12x + 12y = 2
To find the point on the line that is closest to the origin, we need to minimize the distance between the origin and a generic point (x, y) on the line. The distance between two points is given by the Euclidean distance formula:
d = sqrt(x^2 + y^2)
Now, to minimize this distance, we can minimize the square of the distance. So, instead of minimizing d, we will minimize d^2:
D = x^2 + y^2
We have two functions: g(x, y) = 12x + 12y - 2 (the equation of the line) and f(x, y) = x^2 + y^2 (the square of the distance from the origin).
Now, we set up the Lagrange multiplier equation:
∇f = λ∇g
Let's compute the gradients:
∇f = <2x, 2y>
∇g = <12, 12>
Equating these two gradients, we get:
<2x, 2y> = λ<12, 12>
Now, let's break it down into two separate equations:
2x = 12λ (equation 1)
2y = 12λ (equation 2)
To eliminate the λ from these equations, divide equation 1 by 12 and equation 2 by 12:
x/6 = y/6
From this equation, we can conclude that x = y (equation 3).
Now, substitute x = y (equation 3) into the equation of the line (12x + 12y = 2):
12(x) + 12(x) = 2
24x = 2
x = 2/24
x = 1/12
Using equation 3, we find that y = 1/12 as well.
Therefore, the point (x0, y0) closest to the origin on the line 12x + 12y = 2 is given by (1/12, 1/12).
It's important to double-check each step and calculation to ensure accuracy. If your online homework is showing your answer as incorrect, please make sure to re-evaluate the calculations and verify if there were any transcription or arithmetic errors that might have led to an incorrect result.