# Differential Calculus Optimization Problem?

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 find 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.

### 2 Answers

- ThomasLv 74 years agoFavorite Answer
f=xy+z(x+y)

pf/px=y+z, pf/py=x+z

The two partial derivatives can only be equal when x=y....

since z=10-x-y we can say z=10-2x so the original equation can be expressed...

f=x^2+x(10-2x)+x(10-2x)

f=x^2+2x(10-2x)

f=x^2+20x-4x^2

f=20x-3x^2

df/dx=20-6x and since d2f/dx2=-6, when df/dx=0 it is an absolute maximum for f(x)...

df/dx=0 only when 6x=20, x=10/3

(so x=y=10/3, z=10/3, then x=y=z=10/3)

Since f(x)=20x-3x^2 the maximum value is:

f(10/3)=100/3

if you look back at the original equation...

3(10/3)^2=300/9=100/3

- az_lenderLv 74 years ago
When one of the three numbers is nearly 0, the maximum is 25. When one of the three numbers is 1, the maximum is 20.25+4.5+4.5 = 29.25. When one of the three numbers is 2, the maximum is 16 + 8 + 8 = 32. When one of the three numbers is 3, the maximum is 12.25 + 10.5 + 10.5 = 33.25. When one of the three numbers is 10/3, the maximum is 3*(10/3)^2 = 33.3333... . When one of the three numbers is 4, the maximum is 9 + 12 + 12 = 33. It is plain that the maximum is 100/3, but I'll do something more like a proof:

The function representing the sum of the three products can be seen as a function of z only, viz.,

[(10-z)/2]^2 + 2z(10-z)

= 25 - 5z + z^2/4 + 20z - 2z^2

= 25 + 15z - 2.25 z^2.

The derivative of this function, with respect to z, is

15 - 4.5z, which is 0 when z = 15/4.5 = 10/3.

Then the max value is

(10/3)^2 + (10/3)^2 + (10/3)^2 = 100/3.