Cost Function Feb558
1. **Stating the problem:**
We are given the cost function $$C(x,y) = e^y (32x - 16x^2 - 7) + 2y e^{-x^2 - y^2}$$ where $x$ and $y$ represent production quantities of components.
2. **Understanding the function:**
This function combines exponential terms and polynomial expressions in $x$ and $y$. We might be interested in analyzing its behavior, finding critical points, or understanding cost implications.
3. **Formula and rules:**
- The exponential function $e^t$ is always positive.
- Polynomial terms can be positive or negative depending on $x$ and $y$.
- To find critical points, we would compute partial derivatives $\frac{\partial C}{\partial x}$ and $\frac{\partial C}{\partial y}$ and set them to zero.
4. **Partial derivatives:**
$$\frac{\partial C}{\partial x} = e^y (32 - 32x) + 2y e^{-x^2 - y^2} (-2x) = e^y (32 - 32x) - 4xy e^{-x^2 - y^2}$$
$$\frac{\partial C}{\partial y} = e^y (32x - 16x^2 - 7) + 2 e^{-x^2 - y^2} + 2y e^{-x^2 - y^2} (-2y) = e^y (32x - 16x^2 - 7) + 2 e^{-x^2 - y^2} - 4 y^2 e^{-x^2 - y^2}$$
5. **Setting derivatives to zero for critical points:**
Solve the system:
$$e^y (32 - 32x) - 4xy e^{-x^2 - y^2} = 0$$
$$e^y (32x - 16x^2 - 7) + 2 e^{-x^2 - y^2} - 4 y^2 e^{-x^2 - y^2} = 0$$
6. **Interpretation:**
These equations are transcendental and may require numerical methods to solve.
**Final answer:**
The cost function is $$C(x,y) = e^y (32x - 16x^2 - 7) + 2y e^{-x^2 - y^2}$$ with partial derivatives as above for analysis of critical points.