Answer:
[tex](9/2, 3/2)[/tex]
Step-by-step explanation:
We have the equation:
[tex]3x+y=15[/tex]
Which is equivalent to 15 among all pairs of (x, y).
We want to find the pair of solutions (x, y) such that:
[tex]x^2+y^2[/tex]
Is minimum.
Note that our given equation is a line.
And the equation x²+y² is a circle centered on the origin.
In other words, we want to find the radius of the circle such that it is tangent to our line at 3x+y=15.
It must be tangent because this guarantees that it is the smallest value of x²+y².
It's good if we have a visual of this. I've graphed the given linear equation. Please refer to it.
If you remember in geometry, in order for the radius to be tangent to a line, the radius must be perpendicular to our line.
So, let's find the perpendicular equation to our line. Our original equation is:
[tex]3x+y=15[/tex]
Subtract 3x from both sides:
[tex]y=-3x+15[/tex]
So, the slope of our original equation is -3.
This means that the slope of our perpendicular line must be the negative reciprocal of -3. Namely, it is 1/3.
And since our circle is centered on the origin, this line will go through the origin. Therefore, our perpendicular equation is:
[tex]y=\frac{1}{3}x[/tex]
Graphing this will yield (please refer to the second graph):
Therefore, the intersection between our old and new line is at (4.5, 1.5).
Therefore, the (x, y) value that grants the minimum sum is 9/2 and 3/2.
We can check this by substituting them into our second equation. This yields:
[tex](9/2)^2+(5/2)^2[/tex]
Square and add:
[tex]=81/4+9/4=22.5[/tex]
Note that 22.5 is the radius squared.
Graphing this gives us (please refer to the third graph):
We can see that it is indeed tangent to our line. And it is the lowest value of P that does so.
So, our answer is x=9/2 and y=3/2.
