5 - Optimization & constraints

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Calculus of Multivariable Functions

Section 5 Use of the Partial Derivatives: Optimization of Functions Subject to the Constraints

Practise questions

1. Using the Lagrange method find the critical points of the function where the function may be optimized:

$$f(x,y) = 6x^{2} - 2xy + 4y^{2} \qquad\qquad \text{Subject to } g(x,y) = x + y = 36$$

Write the Lagrange function: $$L(x,y,\mu) \equiv \color{red}{f(x,y)} - \mu (\color{purple}{g(x,y) - k})$$ $$L(x,y,\mu) = 6x^{2} - 2xy + 4y^{2} - \mu (x + y - 36)$$ Set each first-order partial derivatives to be zero: $$\frac{\partial L}{\partial x} = 12x - 2y - \mu = 0 \qquad\qquad\qquad \text{(1)}$$ $$\frac{\partial L}{\partial y} = -2x + 8y - \mu = 0 \qquad\qquad\quad\; \text{(2)}$$ $$\frac{\partial L}{\partial \mu } = -(x + y - 36) = 0 \qquad\qquad\quad \text{(3)}$$ From equations (1) and (2): $$y = 1.4x$$ Use $y = 1.4x$ in equation (3) $$2.4x = 36$$ $$x = 15$$ $$y = 1.4x = 21$$ At $(15, 21)$ the objective function may be optimized subject to the given constraint.

2. Using the Lagrange method find the critical points of the function where the function may be optimized:

$$f(x,y) = 6x^{2} - 4xy + 4y^{2} \qquad\qquad \text{Subject to } g(x,y) = x + 2y = 36$$

Write the Lagrange function: $$L(x,y,\mu) \equiv \color{red}{f(x,y)} - \mu (\color{purple}{g(x,y) - k})$$ $$L(x,y,\mu) = 6x^{2} - 4xy + 4y^{2} - \mu (x + 2y - 36)$$ Set each first-order partial derivatives to be zero: $$\frac{\partial L}{\partial x} = 12x - 4y - \mu = 0 \qquad\qquad\qquad \text{(1)}$$ $$\frac{\partial L}{\partial y} = -4x + 8y - 2\mu = 0 \qquad\qquad\quad\; \text{(2)}$$ $$\frac{\partial L}{\partial \mu } = -(x + 2y - 36) = 0 \qquad\qquad\quad \text{(3)}$$ Divide equation (2) by $2$: $$-2x + 4y - \mu = 0 \qquad\qquad\qquad\qquad\quad \text{(4)}$$ From equations (1) and (4): $$14x = 8y$$ $$y = 1.75x$$ Use $y = 1.75x$ in equation (3) $$4.5x = 36$$ $$x = 8$$ $$y = 1.75(x) = 1.75(8) = 14$$ At $(8, 14)$ the objective function may be optimized subject to the given constraint.

3. Using the geometric approach find the critical points of the function where the function may be optimized:

$$f(x,y) = x^{0.5}y^{0.5} \qquad\qquad \text{Subject to } g(x,y) = x + 2y = 60$$

From the objective function $f(x,y) = x^{0.5}y^{0.5}$ find $\frac{f_{x}}{f_{y}}$: $$\frac{f_{x}}{f_{y}} = \frac{y}{x}$$ From the constraint function $g(x,y) = x + 2y = 60$ find $\frac{g_{x}}{g_{y}}$: $$\frac{g_{x}}{g_{y}} = \frac{1}{2}$$ Find the relation using $\frac{f_{x}}{f_{y}} = \frac{g_{x}}{g_{y}}$: $$\frac{y}{x} = \frac{1}{2}$$ $$x = 2y$$ Use the relation obtained, $x = 2y$ in the constraint function: $$x + 2y = 60$$ $$2y + 2y = 60$$ $$4y = 60$$ $$y = 15, \qquad x = 30$$ At $(30, 15)$ the objective function may be optimized subject to the given constraint.

4. Using the geometric approach find the critical points of the function where the function may be optimized:

$$f(x,y) = x^{0.6}y^{0.4} \qquad\qquad \text{Subject to } g(x,y) = x + y = 60$$

From the objective function $f(x,y) = x^{0.6}y^{0.4}$ find $\frac{f_{x}}{f_{y}}$: $$\frac{f_{x}}{f_{y}} = \frac{3y}{2x}$$ From the constraint function $g(x,y) = x + y = 60$ find $\frac{g_{x}}{g_{y}}$: $$\frac{g_{x}}{g_{y}} = 1$$ Find the relation using $\frac{f_{x}}{f_{y}} = \frac{g_{x}}{g_{y}}$: $$\frac{3y}{2x} = 1$$ $$x = 1.5y$$ Use the relation obtained, $x = 1.5y$ in the constraint function: $$x + y = 60$$ $$1.5y + y = 60$$ $$2.5y = 60$$ $$y = 24, \qquad x = 36$$ At $(36, 24)$ the objective function may be optimized subject to the given constraint.