Question

Use Lagrange multipliers to find three numbers whose sum is 30 and the product P = x3y4z is a maximum. Choose the answer for the smallest of the three values. Question 20 options: a) 21/4 b) 5 c) 15/4 d) 3

Answer 1

We want to maximize
`x^3y^4z` subject to the constraint
`x+y+z=30`.

The Lagrangian is

`L(x,y,z,\lambda)=x^3y^4z-\lambda(x+y+z-30)`

with critical points where the derivatives vanish:

`L_x=3x^2y^4z-\lambda=0`

`L_y=4x^3y^3z-\lambda=0`

`L_z=x^3y^4-\lambda=0`

`L_\lambda=x+y+z-30=0`

`\implies\lambda=3x^2y^4z=4x^3y^3z=x^3y^4`

We have

`3x^2y^4z-4x^3y^3z=x^2y^3z(3y-4x)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}\\z=0,\text{ or}\\3y=4x\end{cases}`

`3x^2y^4z-x^3y^4=x^2y^4(3z-x)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}\\3z=x\end{cases}`

`4x^3y^3z-x^3y^4=x^3y^3(4z-y)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}4z=y\end{cases}`

Let's work with
`x=3z` and
`y=4z`, for which we have

`x+y+z=8z=30\implies z=\frac{15}4\implies\begin{cases}x=\frac{45}4\\y=15\end{cases}`

The smallest of these is C. 15/4.