Question

maximum value of x3y3 + 3 x*y when x+y = 8

Answer 1

i think this will be maximum when xy is a maximum

To find x when xy is maimum:

let z = xy

x + y = 8
y = 8-x
z = x(8 - x)
z' = 8 - 2x
x = 4

xy = 4*4 = 16


maximum value is 4^3*4^3 + 3*4*4
= 4144

Answer 2

* z' = 8 - 2x

for max/min
8 - 2x = 0
giving x = 4

maximum value as second derivative is -2 (negative0

Answer 3

Another method, using Lagrange multipliers. We want to maximize \(x^3y^3+3xy\) subject to \(x+y=8\).

Basically we let \(f(x,y)=x^3y^3+3xy\) and \(g(x,y)=x+y=8\), and we have
\[\nabla f(x,y)=\lambda \nabla g(x,y)\]
which gives the system
\[\begin{cases}
3x^2y^3+3y=\lambda(1+0)=\lambda\\
3x^3y^2+3x=\lambda(0+1)=\lambda\\
x+y=8
\end{cases}\]
Setting the first two equations equal to each other:
\[\begin{align*}3x^2y^3+3y&=3x^3y^2+3x\\
x^2y^3+y&=x^3y^2+x\\
y\left(x^2y^2+1\right)&=x\left(x^2y^2+1\right)\\
y&=x\end{align*}\]
Substituting into the second equation, \(x+y=2x=8\) gives \(x=4\), and hence \(y=4\).