Suppose that \(f(x, y)=x y .\) Find the maximum value for / if \(x\) and \(y\) are constrained to sum to \(1 .\) Solve this problem in two ways: by substitution and by using the Langrangian multiplier method.

Since we know that \(x+y=1\), we can easily solve for one variable in terms of the other. Let's solve for \(y\): $$y = 1-x$$ Now, we can substitute this expression for \(y\) in the function \(f(x, y)\): $$f(x, 1-x) = x(1-x)$$ To find the maximum value, we should find the critical points by taking the derivative of the function with respect to \(x\) and set it to zero: $$\frac{d}{dx}(x(1-x)) = (1-x)-x = 1-2x$$ Now, let's find the critical points: $$1-2x=0 \Rightarrow x=\frac{1}{2}$$ Substitute back into \(y\) expression: $$y=1-x \Rightarrow y=1-\frac{1}{2}=\frac{1}{2}$$ The maximum value for \(f(x,y)\) is at the point \((\frac{1}{2},\frac{1}{2})\), so: $$f_{max}(x, y) = f\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{1}{2} * \frac{1}{2} = \frac{1}{4}$$

To solve the problem by using the Lagrangian multiplier method, we first set up the Lagrangian function with a multiplier \(\lambda\), let's denote \(g(x,y)=x+y\): $$\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda(g(x,y)-1) = xy - \lambda(x+y-1)$$ Now, in order to find the maximum value, we have to find where the gradient of the Lagrangian is equal to zero: $$ \nabla \mathcal{L}(x, y, \lambda) = \begin{bmatrix} \frac{\partial \mathcal{L}}{\partial x}\\ \frac{\partial \mathcal{L}}{\partial y}\\ \frac{\partial \mathcal{L}}{\partial \lambda} \end{bmatrix} = \begin{bmatrix} y - \lambda \\ x - \lambda \\ x + y - 1 \end{bmatrix} = 0$$ Solve these equations simultaneously for \(x\), \(y\), and \(\lambda\): $$\begin{cases} y - \lambda = 0 \\ x - \lambda = 0 \\ x + y - 1 = 0 \end{cases}$$ From the first two equations, we can conclude that \(x=y\), and the third equation becomes: $$x+x-1=0 \Rightarrow x=\frac{1}{2}$$ Substitute back into the first equation: $$y - \lambda = 0 \Rightarrow y=\lambda=\frac{1}{2}$$ The maximum value for \(f(x,y)\) is obtained at the point \((\frac{1}{2},\frac{1}{2})\), so: $$f_{max}(x, y) = f\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{1}{2} * \frac{1}{2} = \frac{1}{4}$$ Both methods yield the same maximum value of \(\frac{1}{4}\) at the point \((\frac{1}{2},\frac{1}{2})\).