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Solving for profit function $\pi (w,p)$ given the output production function $f(z) = \sqrt{2z_1 + 3z_2}$.

I approached this problem by trying to solve the $p\nabla f(z) = w$. This is derived from setting up the Lagrangian for the Profit Maximization Problem, \begin{align*} \text{maximize } &pf(z)-w^Tz\\\ \Rightarrow \mathcal{L}(z) &= pf(z) -w^Tz \end{align*} Then taking the partial of the Lagrange to zero, \begin{align*} \frac{\partial \mathcal{L}}{\partial z} = 0 = p\nabla f(z) - w\\ \Rightarrow p\nabla f(z) = w. \end{align*} The issue is, I thought that I could solve for an optimal $z^*$, but that does not seem possible, but I know that a solution exists.

To showing this issue simply, let $q=f(z)$, then the gradient is: \begin{align} \nabla f(z) = \begin{bmatrix} 1/q\\ 3/2q \end{bmatrix} \end{align} So solving our equation $p\nabla f(z) = w$, should let us solve for $z_1,z_2$, but as you can see, \begin{align*} \begin{bmatrix} 1/q\\ 3/2q \end{bmatrix} = \begin{bmatrix} w_1/p\\ w_2/p \end{bmatrix}\\ \Rightarrow \begin{bmatrix} q\\ q \end{bmatrix} = \begin{bmatrix} p/w_1\\ 3p/2w_2 \end{bmatrix}\\ \Rightarrow \begin{bmatrix} \sqrt{2z_1 + 3z_2}\\ \sqrt{2z_1 + 3z_2} \end{bmatrix} = \begin{bmatrix} p/w_1\\ 3p/2w_2 \end{bmatrix} \end{align*} This shows I cannot isolate $z_1$ or $z_2$. Without an optimal $z^*=<z_1^*,z_2^*>$, I cannot find my profit function $\pi(w,p) = pf(z^*) - w^Tz^*$.

EDIT My guess is that actually, for some output $q$, my profit function is what I already solved for, $\pi(w,p)=\max{\{pw_1, 3p/2w_2\}}$

Can anyone confirm this?

Solving for profit function $\pi (w,p)$ given the output production function $f(z) = \sqrt{2z_1 + 3z_2}$.

I approached this problem by trying to solve the $p\nabla f(z) = w$. This is derived from setting up the Lagrangian for the Profit Maximization Problem, \begin{align*} \text{maximize } &pf(z)-w^Tz\\\ \Rightarrow \mathcal{L}(z) &= pf(z) -w^Tz \end{align*} Then taking the partial of the Lagrange to zero, \begin{align*} \frac{\partial \mathcal{L}}{\partial z} = 0 = p\nabla f(z) - w\\ \Rightarrow p\nabla f(z) = w. \end{align*} The issue is, I thought that I could solve for an optimal $z^*$, but that does not seem possible, but I know that a solution exists.

To showing this issue simply, let $q=f(z)$, then the gradient is: \begin{align} \nabla f(z) = \begin{bmatrix} 1/q\\ 3/2q \end{bmatrix} \end{align} So solving our equation $p\nabla f(z) = w$, should let us solve for $z_1,z_2$, but as you can see, \begin{align*} \begin{bmatrix} 1/q\\ 3/2q \end{bmatrix} = \begin{bmatrix} w_1/p\\ w_2/p \end{bmatrix}\\ \Rightarrow \begin{bmatrix} q\\ q \end{bmatrix} = \begin{bmatrix} p/w_1\\ 3p/2w_2 \end{bmatrix}\\ \Rightarrow \begin{bmatrix} \sqrt{2z_1 + 3z_2}\\ \sqrt{2z_1 + 3z_2} \end{bmatrix} = \begin{bmatrix} p/w_1\\ 3p/2w_2 \end{bmatrix} \end{align*} This shows I cannot isolate $z_1$ or $z_2$. Without an optimal $z^*=<z_1^*,z_2^*>$, I cannot find my profit function $\pi(w,p) = pf(z^*) - w^Tz^*$.

Solving for profit function $\pi (w,p)$ given the output production function $f(z) = \sqrt{2z_1 + 3z_2}$.

I approached this problem by trying to solve the $p\nabla f(z) = w$. This is derived from setting up the Lagrangian for the Profit Maximization Problem, \begin{align*} \text{maximize } &pf(z)-w^Tz\\\ \Rightarrow \mathcal{L}(z) &= pf(z) -w^Tz \end{align*} Then taking the partial of the Lagrange to zero, \begin{align*} \frac{\partial \mathcal{L}}{\partial z} = 0 = p\nabla f(z) - w\\ \Rightarrow p\nabla f(z) = w. \end{align*} The issue is, I thought that I could solve for an optimal $z^*$, but that does not seem possible, but I know that a solution exists.

To showing this issue simply, let $q=f(z)$, then the gradient is: \begin{align} \nabla f(z) = \begin{bmatrix} 1/q\\ 3/2q \end{bmatrix} \end{align} So solving our equation $p\nabla f(z) = w$, should let us solve for $z_1,z_2$, but as you can see, \begin{align*} \begin{bmatrix} 1/q\\ 3/2q \end{bmatrix} = \begin{bmatrix} w_1/p\\ w_2/p \end{bmatrix}\\ \Rightarrow \begin{bmatrix} q\\ q \end{bmatrix} = \begin{bmatrix} p/w_1\\ 3p/2w_2 \end{bmatrix}\\ \Rightarrow \begin{bmatrix} \sqrt{2z_1 + 3z_2}\\ \sqrt{2z_1 + 3z_2} \end{bmatrix} = \begin{bmatrix} p/w_1\\ 3p/2w_2 \end{bmatrix} \end{align*} This shows I cannot isolate $z_1$ or $z_2$. Without an optimal $z^*=<z_1^*,z_2^*>$, I cannot find my profit function $\pi(w,p) = pf(z^*) - w^Tz^*$.

EDIT My guess is that actually, for some output $q$, my profit function is what I already solved for, $\pi(w,p)=\max{\{pw_1, 3p/2w_2\}}$

Can anyone confirm this?

Source Link
Tsangares
Tsangares
  • 145
  • 4

Solving for profit function $\pi (w,p)$ given the output of production function $f(z) = \sqrt{2z_1 + 3z_2}$

Solving for profit function $\pi (w,p)$ given the output production function $f(z) = \sqrt{2z_1 + 3z_2}$.

I approached this problem by trying to solve the $p\nabla f(z) = w$. This is derived from setting up the Lagrangian for the Profit Maximization Problem, \begin{align*} \text{maximize } &pf(z)-w^Tz\\\ \Rightarrow \mathcal{L}(z) &= pf(z) -w^Tz \end{align*} Then taking the partial of the Lagrange to zero, \begin{align*} \frac{\partial \mathcal{L}}{\partial z} = 0 = p\nabla f(z) - w\\ \Rightarrow p\nabla f(z) = w. \end{align*} The issue is, I thought that I could solve for an optimal $z^*$, but that does not seem possible, but I know that a solution exists.

To showing this issue simply, let $q=f(z)$, then the gradient is: \begin{align} \nabla f(z) = \begin{bmatrix} 1/q\\ 3/2q \end{bmatrix} \end{align} So solving our equation $p\nabla f(z) = w$, should let us solve for $z_1,z_2$, but as you can see, \begin{align*} \begin{bmatrix} 1/q\\ 3/2q \end{bmatrix} = \begin{bmatrix} w_1/p\\ w_2/p \end{bmatrix}\\ \Rightarrow \begin{bmatrix} q\\ q \end{bmatrix} = \begin{bmatrix} p/w_1\\ 3p/2w_2 \end{bmatrix}\\ \Rightarrow \begin{bmatrix} \sqrt{2z_1 + 3z_2}\\ \sqrt{2z_1 + 3z_2} \end{bmatrix} = \begin{bmatrix} p/w_1\\ 3p/2w_2 \end{bmatrix} \end{align*} This shows I cannot isolate $z_1$ or $z_2$. Without an optimal $z^*=<z_1^*,z_2^*>$, I cannot find my profit function $\pi(w,p) = pf(z^*) - w^Tz^*$.