Suppose x; y, and z are positive numbers that sum to 10. What is the largest possible value of xy + xz + yz?
(a) First, suppose that z is a �fixed parameter. Then we have to �find nonnegative numbers x and y (depending on the �fixed value z such that x+y = 10-z), such that xy + xz + yz is maximized.
(b) To fi�nd the global maximum, plug in the values for x and y obtained in (a) in
order to rewrite xy + xz + yz as a function of a single variable. Then maximize
this function.
since x+y+z = 10, we want the max of
F(x,y) = xy + (x+y)(10-x-y)
Fx = -2x-y+10
Fy = -x-2y+10
Fx = Fy = 0 at (10/3,10/3)
So, max of F(x,y) = 100/3
That is, when x,y,z are all equal, at 10/3.
Using common sense:
for any value xy, where x + y is a constant,
the largest value xy is obtained when xy is a perfect square,
So the largest value of xy+xz+zy is obtained when all three are the same value
but 10 cannot be split into 3 equal parts
so the best we can do is split them into
3,3, and 4
so largest sum = 3x3 + 3x4 + 3x4 = 33
confirmation: - a little program
5 big = 0
10 for x = 1 to 10
20 for y = 1 to 10
30 for z = 1 to 10
40 sum = x*y+x*z+y*z
45 if x+y+z > 10 then 60
50 if sum > big then big = sum : print x,y,z,sum
60 next z
70 next y
80 next x
>run
1 1 1 3
1 1 2 5
1 1 3 7
1 1 4 9
1 1 5 11
1 1 6 13
1 1 7 15
1 1 8 17
1 2 6 20
1 2 7 23
1 3 6 27
1 4 5 29
2 3 5 31
2 4 4 32
3 3 4 33 <--- sum
>
To find the largest possible value of xy + xz + yz, we need to maximize this expression given that x, y, and z are positive numbers that sum to 10.
First, let's consider (a) where z is a fixed parameter. In this case, we need to find non-negative numbers x and y such that x + y = 10 - z. We can rewrite the expression xy + xz + yz as:
xy + xz + yz = x(y + z) + yz = (y + z)(x + z)
Now, we have a quadratic expression in terms of a single variable (x + z) since y + z is a constant. To maximize this quadratic expression, we need to find the vertex of the parabola.
The vertex of a quadratic function in the form ax^2 + bx + c is given by the coordinate (-b/2a, f(-b/2a)), where f(x) is the value of the quadratic function at x.
In our case, the quadratic expression is (y + z)(x + z). So, the value of x that corresponds to the vertex is -z, and the expression reaches its maximum value at x = -z.
Now, let's move to (b). We will use the values for x and y obtained in (a).
Substituting x = 10 - z and y = z, we can rewrite xy + xz + yz as:
(10 - z)(z) + (10 - z)z + z^2 = 10z - z^2 + 10z - z^2 + z^2 = 20z - 2z^2
To maximize this expression, we can take the first derivative with respect to z and set it equal to 0 to find the critical points:
d/dz (20z - 2z^2) = 20 - 4z = 0
Solving this equation, we find z = 5.
To determine if this is a maximum or minimum, we can take the second derivative:
d^2/dz^2 (20z - 2z^2) = -4
Since the second derivative is negative, we have a concave down parabola, and z = 5 corresponds to the maximum value.
Therefore, when x, y, and z are positive numbers that sum to 10, the largest possible value of xy + xz + yz is obtained when z = 5.