(x,y,z)∈R3 are points that lie on the plane x+2y+3z=78, and lie on the sphere x^2+y^2+z^2=468. The maximum value of x has the form a/b, where a and b are coprime positive integers. What is the value of a+b?
To find the maximum value of x, we need to find the intersection points of the plane and the sphere.
Start by expressing x in terms of y and z from the equation of the plane: x = 78 - 2y - 3z.
Substitute this value of x into the equation of the sphere to eliminate x:
(78 - 2y - 3z)^2 + y^2 + z^2 = 468.
Expanding and simplifying, we get:
4y^2 + 4yz + 4z^2 - 312y - 468z + 4344 = 0.
Now, we have a quadratic equation in terms of y and z. To solve it, we can apply the quadratic formula:
y = (-4z ± √(16z^2 - 16(4z^2 - 312z + 4344))) / (8).
Simplifying further, we have:
y = (-4z ± √(16z^2 - 64z^2 + 4992z - 34752)) / (8).
y = (-4z ± √(-48z^2 + 4992z - 34752)) / (8).
y = (-z ± √(-12z^2 + 1248z - 8694)) / (2).
Now, we substitute the expression for y back into the equation of the plane to get the corresponding values of x:
x = 78 - 2(-z ± √(-12z^2 + 1248z - 8694))/2 - 3z.
Simplifying further, we have:
x = 39 + z ± √(-12z^2 + 1248z - 8694)/2 - 3z.
Combining like terms:
x = 39 - z ± √(-12z^2 + 1248z - 8694)/2.
Now, we can observe that the maximum value of x occurs when the expression inside the square root, -12z^2 + 1248z - 8694, is minimized. To find the minimum value of this quadratic expression, we can find its vertex.
The x-coordinate of the vertex is given by: z = -b/(2a), where a = -12 and b = 1248.
Plugging in these values, we get:
z = -1248/(2*(-12)) = 104.
Substituting z = 104 back into the expression for x, we get:
x = 39 - 104 ± √(-12(104)^2 + 1248(104) - 8694)/2.
Simplifying further:
x = 39 - 104 ± √(-12(10816) + 129792 - 8694)/2.
x = 39 - 104 ± √(-129792 + 129792 - 8694)/2.
x = 39 - 104 ± 0/2.
x = 39 - 104 ± 0.
Therefore, the maximum value of x is 39.
Since 39 is already an integer, we can rewrite it as a fraction with a denominator of 1: 39/1.
Therefore, the value of a+b is 39+1 = 40.