Sum and Square Sum

If $x = 0,$ we have $2y + 3z = 78$ and $y^2 + z^2 = 468.$

Find y so we have, $y = (-3/2)*z + 39$

Subsitute $y$ into $y^2 + z^2 = 468,$ you'll get $(9/4)z^2 - 117z + 1521 + z^2 = 468 \Rightarrow 13z^2 - 468z + 4212 = 0$ $\Rightarrow z^2 - 36*z + 324 = 0 \Rightarrow (z - 18)^2 = 0$

which results to z = 18, and y = 39 - (3/2)*18 = 12.

Therefore, the circle that is the intersection of the given plane and sphere is tangent to the yz-plane at $(0, 12, 18)$ . So, there is a maximum value of at the point at the opposite end of the diameter to the point $(0, 12, 18).$

To find this point, we need to find the center of the circle of intersection. This is the point on the plane where the normal line to the plane passing through this point passes through the center of the sphere, i.e., the origin. The normal vector to the plane is $<1,2,3>$ , so let the point on the plane corresponding to the center of the circle of intersection be $(a,b,c)$ , we have the equations $x = a + t, y = b + 2t$ and $z = c + 3t$ such that $a + 2b + 3c = 78$ .

To have the line pass through the origin, $a = -t, b = -2t$ and $c = -3t$ , therefore, $-t - 4t - 9t = 78 \Rightarrow -14t = 78 \Rightarrow t = -39/7.$

So, the center of the circle of intersection is $(39/7, 78/7, 117/7).$

The center of the circle is the midpoint of $(0,12,18)$ and the point $(X,Y,Z)$ that we are looking for. So, $(X + 0)/2 = 39/7 \Rightarrow X = 78/7$ .

(also $(Y + 12)/2 = 78/7 \Rightarrow Y = 72/7,$ and $(Z + 18)/2 = 117/7 \Rightarrow Z = 108/7.$

Since we now have the maximum X = a/b where a and b are co-prime, the value of $a + b$ is $78 + 7$ which is equal to $85.$

According to the topic, we have: $x^2+y^2+(\frac {78-x-2y}{3})^2=468$ $\Leftrightarrow 13y^2+2(2x-156)y+10x^2-156x+1872=0$

$y$ exists so $(2x-156)^2-13(10x^2-156x+1872)\geq0$ $\Leftrightarrow 7x^2-78x \leq 0$ $\Leftrightarrow 0 \leq x \leq \frac {78}{7}$

Thus, $a+b=85$ .

We see that we can eliminate either y or z from the second equation and do so; choosing to eliminate y by substituting $y=\frac{78-3z-x}{2}$ into $x^{2}+y^{2}+z^{2}=468$ gives $x^{2}+\frac{1}{4}(2^{2}.3^{2}.13^{2}+3^{2}.z^{2}+x^{2}-2^{2}.3^{2}.13z-2^{2}.3.13x+2.3xz)+z^{2}=2^{2}.3^{2}.13 \\$ $5x^{2}+2^{2}.3^{2}.13^{2}-2^{2}.3^{2}.13z-2^{2}.3.13x+2.3xz+13z^{2}=2^{4}.3^{2}.13$

We will write this as a quadratic in z giving $13z^{2}+z(2.3.x-2^{2}.3^{2}.13)+2^{2}.3^{2}.13^{2}-2^{4}.3^{2}.13-2^{2}.3.13x+5x^{2}=0$ For this to have real solutions, the discriminant of this must be $\ge 0$ . For it to have one solution, which MUST correspond with an extremum for x, we get $(2.3.x-2^{2}.3^{2}.13)^{2}-2^{2}.13(2^{2}.3^{2}.13^{2}-2^{4}.3^{2}.13-2^{2}.3.13x+5x^{2})=0$ Then $-2^{5}.7x^{2}+2^{4}.3.13x(13-3^{2})+2^{4}.3^{2}.13^{2}(3^{2}-13+2^{2})=0$ $-7x^{2}+2.3.13x=0$ Which gives $x=0$ and $x=78/7$ as the two extrema, the latter of these is therefore the maximum value of $x$

Consider the 13 integers $a_1=a_2=a_3=a_4=\frac{y}{2}$ , $a_5=a_6=\ldots=a_{13}=\frac{z}{3}$ . By using Cauchy-Schwarz Inequality, we have $(\sum_{i=1}^{13}a_i^2)(\sum_{i=1}^{13}{1^2}) \geq (\sum_{i=1}^{13}a_i)^2$ $(4(\frac{y}{2})^2+9(\frac{z}{3})^2)(13) \geq (4(\frac{y}{2})+9(\frac{z}{3}))^2$ $(y^2+z^2)(13) \geq (2y+3z)^2$ $(468-x^2)(13) \geq (78-x)^2$ Expanding and rearranging gives us $14x^2-156x \leq 0 \Rightarrow 0 \leq x \leq \frac{156}{14}=\frac{78}{7}$ so the required answer is $78+7=85$

maxima and minima problems can be easily silved by the method of Lagrange Multiplier

now as we have to find out maximum value of x consider the functions

f(x) = x ==> ∇f = <1, 0, 0> ( where ∇f = df/dx, df/dy, df/dz )

g(x) = x + 2y + 3z = 78 ==> ∇g = <1 , 2, 3>

h(x) = x^2 + y^2 + z^2 = 468 ==> ∇h = <2x, 2y, 2z>

∇f = λ ∇g + μ ∇h

<1, 0, 0> = λ <1, 2, 3> + μ <2x, 2y, 2z>

==> x = (1 - λ)/(2μ), y = -λ/μ, z = -3λ/(2μ)

now by substituting these values into g(x) and h(x) respectively, you get two simultaneous equations, solving them gives (λ, μ) = (-6/7, 1/12), (1, -1/12) ==> x = 78/7, 0

So the maximum value of x is 78/7 ==> gcd(78,7) = 1 ==> a = 78, b = 7 ==> a + b = 85

At the maximum value of $x$ , the line $2y+3z = 78-x$ will be tangent to the circle $y^2 + z^2 = 468-x^2$ at one point.

The line has a slope of $\frac{-2}{3}$ , and by deriving with respect to $y$ we can get that the tangent line to the circle has a slople of $-\frac{y}{z}$ at the point $y$ .

Setting these equal to each other we have $z = \frac{3}{2}y$ , and plugging this into our first equation gives $y=12-\frac{2}{13}x$ .

Then, we substitute this all into our third equation:

$x^2 + (12-\frac{2}{13}x)^2+(\frac{3}{2}(12-\frac{2}{13}x))^2=468$ $\frac{14}{13}x^2=12x$ $x=\frac{78}{7}$ , so our final answer is 85.

Given that x+2y+3z=78 x x+y y+z z=468 By using CAUSHY SQUARTZ INEQUALITY, (2y+3z)(2y+3z)<=(2 2+3 3)(y y+z z) but 2y+3z=78-x and y y+z z=468-x x therefore (78-x) (78-x)<=13 (468-x x) then by expanding above expression 7x x-78x<=0 thereforex=0 or 78/7 hence x=78/7 because x=a/b therefore a=78,b=7 then a+b=78+7=85

"by a simple application of calculus" is hand-waving, not a justification. "since we can assume a not equal to 0" is unclear

Solution 1: We set $x_1 = 2, x_2 = 3, y_1 = y, y_2 = z$ , in the Cauchy-Schwarz inequality. This gives us $( 4+9)( y^2 + z^2 ) \geq (2y+3z)^2$ . Using the 2 equations in the challenge yields $13 ( 468 -x^2 ) \geq ( 78 - x )^2$ . Expanding and simplifying gives $0 \geq 14 x^2 - 156 x$ , which has the solution set $0 \leq x \leq \frac {156}{14} = \frac {78}{7}$ .

We still have to determine if, for $x = \frac {78}{7}$ , there exists values $y, z$ that satisfy the given conditions. In the Cauchy-Schwarz inequality, we have equality if $y=2k, z=3k$ . Adding the condition $x + 2y + 3z = 78$ , this gives $k = \frac {78\cdot 6}{13 \cdot 7} = \frac {6\cdot 6 }{7}$ . In the quadratic polynomial inequality involving $x$ , we have equality as $x = \frac {78}{7}$ . As such, this set of values $(x, y, z) = ( \frac {78}{7}, \frac {72}{7}, \frac {108}{ 7} )$ would satisfy $13(y^2 + z^2) = (2y+3z)^2 = (78 - x)^2 = 13(468-x^2)$ . Here, the first equality follows from equality in Cauchy Schwarz, the second equality follows from the condition $x+2y+3z=78$ and the third equality follows from the quadratic inequality. Hence, $x^2+y^2+z^2 = 468$ .

Thus, the largest value of $x$ is $\frac {78}{7}$ , which gives $a + b = 78 + 7 = 85$ .

Solution 2: First of all, we can express $x$ as $78-2y-3z$ . If we denote $2y+3z$ by $u$ and $3y-2z$ by $v,$ then $y^2+z^2=\frac{1}{13}(u^2+v^2)$ and there is a one-to-one correspondence between pairs $(y,z)$ and $(u,v).$ So we are looking for the minimum of $78-u$ under the constraint $(78-u)^2+\frac{1}{13}(u^2+v^2)=468.$

Rewriting this in terms of $x$ and $v,$ gives $x^2+\frac{1}{13}(78-x)^2+\frac{1}{13}v^2=468.$ Multiplying out, we get $\frac{14}{13}x^2-12x+468+\frac{1}{13}v^2=468,$ which simplifies to $14x^2-156x+v^2=0.$ Since $v$ is an arbitrary real number, $v^2$ is an arbitrary non-negative real number, so this condition is equivalent to $14x^2-156x\leq 0.$ Because the roots of this quadratic equation are $0$ and $\frac{78}{7},$ the largest possible value of $x$ is $\frac{78}{7}.$ Therefore, the answer is $85.$

We have, 2y+3z=78-x and y²+z²=468-x². Now we will apply Cautchy-Swartz inequality[( $\sum_{i=1}^n$ $a_{i}$ ²)( $\sum_{i=1}^n$ $b_{i}$ ²)>=( $\sum_{i=1}^n$ $a_{i}$ $b_{i}$ )²] (2²+3²)(y²+z²)>=(2y+3z)² => 13(468-x²)>=(78-x)² => 14x²-156x<=0 => x€[0,78/7].