The maximum value of the expression \( st + tu + uv \) given the constraint that \( s + t + u + v = 63 \) is achieved when the values of \( s, t, u, \) and \( v \) are optimally distributed. The discussion highlights the importance of balancing the variables to maximize the product terms. The specific configuration that yields the highest value involves setting \( s, t, u, \) and \( v \) to values that are as close to each other as possible, adhering to the constraint of their sum being 63.
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MarkFL said:My solution:
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 said:If $s,\,t,\,u,\,v$ are positive integers with sum $63$, find the maximum value of $st+tu+uv$.
kaliprasad said:$st+tu+uv = (s+u)(v+t) -vs$
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