7 - Constrained optimization

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Calculus of Multivariable Functions

Section 7 Use of Partial Derivatives in Economics; Constrained Optimization

Practise questions

1. Using the Lagrange method, find the critical points of the following constrained optimization if any:

Optimize $Q(x,y) = x^{0.6}y^{0.3}$
subject to $g(x,y) = 4x + 2y = 300$

Form the Lagrange function: $$L(x,y,\mu ) \equiv \color{red}{Q(x,y)} - \mu (\color{purple}{g(x,y) - k}) \quad\;\;$$ $$L(x,y,\mu ) \equiv x^{0.6}y^{0.3} - \mu (4x + 2y - 300)$$ Set each first order partial derivative equal to zero: $$\frac{\partial L}{\partial x} = 0.6x^{-0.4}y^{0.3} - 4\mu = 0 \qquad\qquad\; \text{(1)}$$ $$\frac{\partial L}{\partial y} = 0.3x^{0.6}y^{-0.7} - 2\mu = 0 \qquad\qquad\; \text{(2)}$$ $$\frac{\partial L}{\partial \mu } = -(4x + 2y - 300) = 0 \qquad\qquad \text{(3)}$$ Multiply equation (2) by $2$: $$0.6x^{0.6}y^{-0.7} - 4\mu = 0 \qquad\qquad\qquad\quad\;\; \text{(4)}$$ From equations (1) and (4) we can write: $$0.6x^{-0.4}y^{0.3} = 0.6x^{0.6}y^{-0.7}$$ $$x = y$$ Use $x = y$ in equation (3) to get: $$4x + 2y = 300$$ $$6x = 300 \qquad$$ $$x = 50 \qquad$$ $$y = x = 50$$ The objective function may be optimized at $(50,50)$.

2. Using the geometric approach, find the critical points of the following constrained optimization if any:

Optimize $Q(K,L) = K^{0.6}L^{0.3}$
subject to $g(K,L) = 4K + 2L = 300$

If there exist an interior optimum, then we can apply the following geometric approach to find the point which may be a critical point of the constrained optimization problem:
Find the slope of the objective function: $-\frac{K}{2L}$
Find the slope of the constrained function: $-\frac{2}{4}$
If there exists an interior solution: $$-\frac{K}{2L} = -\frac{2}{4}$$ $$K = L$$ Use $K = L$ in the constraint function to get: $$4K + 2L = 300$$ $$6K = 300 \qquad\;\;$$ $$K = 50 \qquad\;\;$$ $$L = K = 50$$

3. Find the critical points of the following utility function subject to the constraint:

Optimize $u(x,y) = x^{0.5}y^{0.5}$
subject to $u(x,y) = 40x + 20y = 800$

Form the Lagrange function: $$L(x,y,\mu ) \equiv x^{0.5}y^{0.5} - \mu (40x + 20y - 800)$$ Set each first order partial derivative equals zero: $$\frac{\partial L}{\partial x} = 0.5x^{-0.5}y^{0.5} - 40\mu = 0 \qquad\qquad\;\; \text{(1)}$$ $$\frac{\partial L}{\partial y} = 0.5x^{0.5}y^{-0.5} - 20\mu = 0 \qquad\qquad\;\; \text{(2)}$$ $$\frac{\partial L}{\partial \mu } = -(40x + 20y - 800) = 0 \qquad\qquad \text{(3)}$$ Multiply equation (2) by $2$: $$x^{0.5}y^{-0.5} - 40\mu = 0 \qquad\qquad\qquad\qquad\quad\;\; \text{(4)}$$ From equations (1) and (4) we can write: $$0.5x^{-0.5}y^{0.5} = x^{0.5}y^{-0.5}$$ $$\qquad y = 2x$$ Use $y = 2x$ in equation (3) to get: $$40x + 20(2x) = 800$$ $$80x = 800$$ $$x = 10$$ $$y = 2x = 20$$

Note: Geometric approach will also provide the same solution.