MHB Max of $st+tu+uv$ given $s,t,u,v$ sum 63

[anemone]

If $s,\,t,\,u,\,v$ are positive integers with sum $63$, find the maximum value of $st+tu+uv$.

 

[MarkFL]

My solution:

Spoiler

We are given the objective function:

$$f(s,t,u,v)=st+tu+uv$$

Subject to the constraint:

$$g(s,t,u,v)=s+t+u+v-63=0$$

Applying Lagrange multipliers, we obtain the system:

$$t=\lambda$$

$$s+u=\lambda$$

$$t+v=\lambda$$

$$u=\lambda$$

From this, we readily see $t=u$ and then $s=v=0$. Using the constraint, we then find:

$$2t=63\implies t=u=\frac{63}{2}=31.5$$

Since we require the variables to be positive integers, let $(s,t,u,v)=(1,30,31,1)$ (or equivalently $(s,t,u,v)=(1,31,30,1)$) as these permutations are the closest to the real number maximum, then we find:

$$f_{\max}=f(1,30,31,1)=1\cdot30+30\cdot31+31\cdot1=30+930+31=991$$

 

[anemone]

My solution:

Spoiler

We are given the objective function:

$$f(s,t,u,v)=st+tu+uv$$

Subject to the constraint:

$$g(s,t,u,v)=s+t+u+v-63=0$$

Applying Lagrange multipliers, we obtain the system:

$$t=\lambda$$

$$s+u=\lambda$$

$$t+v=\lambda$$

$$u=\lambda$$

From this, we readily see $t=u$ and then $s=v=0$. Using the constraint, we then find:

$$2t=63\implies t=u=\frac{63}{2}=31.5$$

Since we require the variables to be positive integers, let $(s,t,u,v)=(1,30,31,1)$ (or equivalently $(s,t,u,v)=(1,31,30,1)$) as these permutations are the closest to the real number maximum, then we find:

$$f_{\max}=f(1,30,31,1)=1\cdot30+30\cdot31+31\cdot1=30+930+31=991$$

Awesome, MarkFL! :D

 

[kaliprasad]

If $s,\,t,\,u,\,v$ are positive integers with sum $63$, find the maximum value of $st+tu+uv$.

Spoiler

$st+tu+uv = (s+u)(v+t) -vs$
as in the 1st term v and s do not lie in isolation we can maximize $(s+u)(v+t)$ and then minimize $vs$ independently . clearly $(s+u)(v+t)$ is maximum when they are as close as possible
so $s+ u = 32$ and $v+t = 31$ or $s+u = 31$ and $v+ t = 32$
and vs is minimum when $v=s = 1$
so we have $s= 1, u= 31, v= 1, t= 30$ or $s = 1, u = 30, t= 31, v= 1$ and in both cases $st+tu+uv = 31 * 32-1 = 991$ maximum

 

[kaliprasad]

Spoiler

$st+tu+uv = (s+u)(v+t) -vs$
as in the 1st term v and s do not lie in isolation we can maximize $(s+u)(v+t)$ and then minimize $vs$ independently . clearly $(s+u)(v+t)$ is maximum when they are as close as possible
so $s+ u = 32$ and $v+t = 31$ or $s+u = 31$ and $v+ t = 32$
and vs is minimum when $v=s = 1$
so we have $s= 1, u= 31, v= 1, t= 30$ or $s = 1, u = 30, t= 31, v= 1$ and in both cases $st+tu+uv = 31 * 32-1 = 991$ maximum

Spoiler

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