Question

If (𝑥^∗ , 𝑦^∗) is the optimal solution of the problem
maximize 𝑓(𝑥, 𝑦) = 100−𝑒^−𝑥−𝑒 ^−𝑦 
subject to 𝑒𝑥+𝑦=\(\frac{𝑒}{𝑒−1},\) 𝑥 ≥ 0, 𝑦 ≥ 0. 
Then \(\sqrt{\frac{y^*}{x^*}}\) is equal to ________ (round off to 2 decimal places).

Answer 1

Correct Answer: 2.7

Problem: Maximize \[ f(x,y)=100-e^{-x}-e^{-y} \] subject to \[ e^{x}+y=C,\qquad C=\frac{e}{e-1}, \] and \(x\ge0,\;y\ge0\). Let the optimal point be \((x^\ast,y^\ast)\). Find the ratio \(\dfrac{y^\ast}{x^\ast}\) (rounded to two decimals). 

Step 1 — turn maximization into a constrained minimization 
Maximizing \(f(x,y)\) is equivalent to minimizing \[ g(x,y)=e^{-x}+e^{-y} \] subject to the same constraint \(h(x,y)=e^{x}+y-C=0\). 

Step 2 — Lagrange multipliers
Form the Lagrangian \[ \mathcal{L}(x,y,\lambda)=e^{-x}+e^{-y}+\lambda\,(e^{x}+y-C). \] Take partial derivatives and set to zero: \[ \frac{\partial\mathcal{L}}{\partial x}=-e^{-x}+\lambda e^{x}=0 \quad\Rightarrow\quad \lambda=e^{-x}/e^{x}=e^{-2x}, \] \[ \frac{\partial\mathcal{L}}{\partial y}=-e^{-y}+\lambda=0 \quad\Rightarrow\quad \lambda=e^{-y}. \] Equate the two expressions for \(\lambda\): \[ e^{-2x}=e^{-y}\quad\Rightarrow\quad -2x=-y\quad\Rightarrow\quad y=2x. \] So at the optimum \(y^\ast=2x^\ast\), hence \[ \boxed{\frac{y^\ast}{x^\ast}=2.00.} \] 

(Optional check): Substitute \(y=2x\) into the constraint to verify a feasible positive solution: \[ e^{x}+2x=C=\frac{e}{e-1}\approx 1.5819767. \] Solving numerically gives \(x^\ast\approx 0.18773\), \(y^\ast\approx0.37547\), and indeed \(y^\ast/x^\ast\approx2.00\).