Subjects calculus

Cost Function Feb558

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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.