Chapter 15 - Differentiation in Several Variables - 15.8 Lagrange Multipliers: Optimizing with a Constraint - Exercises - Page 833: 39

(a) using the method of Lagrange multipliers we show that $L = {\left( {{h^{2/3}} + {b^{2/3}}} \right)^{3/2}}$ is the minimum length of the ladder. (b) we show that the value of $L$ is also equal to the radius of the circle with center $\left( { - b, - h} \right)$ that is tangent to the graph of $xy = bh$.

(a) From Figure 21, we see that ${L^2} = {\left( {x + b} \right)^2} + {\left( {y + h} \right)^2}$. Let the angle between the ladder and the floor be $\alpha $ when $L$ is minimum length. So, we have $\tan \alpha = \frac{y}{b} = \frac{h}{x}$. Thus, $xy = bh$. Since finding the minimum $L$ is the same as finding the minimum square of $L$, our task is to minimize $f\left( {x,y} \right) = {\left( {x + b} \right)^2} + {\left( {y + h} \right)^2}$ subject to the constraint $g\left( {x,y} \right) = xy - bh = 0$. Step 1. Write out the Lagrange equations Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields $\left( {2\left( {x + b} \right),2\left( {y + h} \right)} \right) = \lambda \left( {y,x} \right)$ So, the Lagrange equations are $2\left( {x + b} \right) = \lambda y$, ${\ \ \ }$ $2\left( {y + h} \right) = \lambda x$ Step 2. Solve for $\lambda$ in terms of $x$ and $y$ Since $\left( {0,0} \right)$ does not satisfy the constraint, we may assume that $x \ne 0$, $y \ne 0$. So, $\lambda \ne 0$. From the results in Step 1, we obtain $\lambda = \frac{{2\left( {x + b} \right)}}{y} = \frac{{2\left( {y + h} \right)}}{x}$ $\frac{{2\left( {x + b} \right)}}{y} = \frac{{2\left( {y + h} \right)}}{x}$ (1) ${\ \ \ \ }$ ${x^2} + bx = {y^2} + hy$ Step 3. Solve for $x$ and $y$ using the constraint Using the constraint $xy - bh = 0$, we get $y = \frac{{bh}}{x}$. Substituting $y = \frac{{bh}}{x}$ in equation (1) gives ${x^2} + bx = \frac{{{b^2}{h^2}}}{{{x^2}}} + \frac{{b{h^2}}}{x}$ ${x^4} + b{x^3} = {b^2}{h^2} + b{h^2}x$ ${x^4} + b{x^3} - b{h^2}x - {b^2}{h^2} = 0$ $\left( {{x^3} - b{h^2}} \right)\left( {x + b} \right) = 0$ So, $x = {\left( {b{h^2}} \right)^{1/3}}$ or $x=-b$. Since $x$ takes on positive value physically, only the solution $x = {\left( {b{h^2}} \right)^{1/3}}$ is applicable. Using $y = \frac{{bh}}{x}$, we obtain the critical point: $\left( {{{\left( {b{h^2}} \right)}^{1/3}},{{\left( {{b^2}h} \right)}^{1/3}}} \right)$. Step 4. Calculate the critical values We evaluate the extreme value at the critical point $\left( {{{\left( {b{h^2}} \right)}^{1/3}},{{\left( {{b^2}h} \right)}^{1/3}}} \right)$: $f\left( {{{\left( {b{h^2}} \right)}^{1/3}},{{\left( {{b^2}h} \right)}^{1/3}}} \right)$ $ = {\left( {{{\left( {b{h^2}} \right)}^{1/3}} + b} \right)^2} + {\left( {{{\left( {{b^2}h} \right)}^{1/3}} + h} \right)^2}$ $ = {\left( {b{h^2}} \right)^{2/3}} + 2b{\left( {b{h^2}} \right)^{1/3}} + {b^2} + {\left( {{b^2}h} \right)^{2/3}} + 2h{\left( {{b^2}h} \right)^{1/3}} + {h^2}$ $ = {\left( {{b^2}{h^4}} \right)^{1/3}} + 2{\left( {{b^4}{h^2}} \right)^{1/3}} + {b^2} + {\left( {{b^4}{h^2}} \right)^{1/3}} + 2{\left( {{b^2}{h^4}} \right)^{1/3}} + {h^2}$ $ = 3{\left( {{b^2}{h^4}} \right)^{1/3}} + 3{\left( {{b^4}{h^2}} \right)^{1/3}} + {b^2} + {h^2}$ $ = {\left( {{h^{2/3}}} \right)^3} + 3{\left( {{b^2}{h^4}} \right)^{1/3}} + 3{\left( {{b^4}{h^2}} \right)^{1/3}} + {\left( {{b^{2/3}}} \right)^3}$ Write $c = {h^{2/3}}$ and $d = {b^{2/3}}$. Since ${\left( {c + d} \right)^3} = {c^3} + 3{c^2}d + 3c{d^2} + {d^3}$, so ${\left( {{h^{2/3}} + {b^{2/3}}} \right)^3} = {\left( {{h^{2/3}}} \right)^3} + 3{h^{4/3}}{b^{2/3}} + 3{h^{2/3}}{b^{4/3}} + {\left( {{b^{2/3}}} \right)^3}$ $ = {\left( {{h^{2/3}}} \right)^3} + 3{\left( {{b^2}{h^4}} \right)^{1/3}} + 3{\left( {{b^4}{h^2}} \right)^{1/3}} + {\left( {{b^{2/3}}} \right)^3}$ Therefore, $f\left( {{{\left( {b{h^2}} \right)}^{1/3}},{{\left( {{b^2}h} \right)}^{1/3}}} \right) = {\left( {{h^{2/3}} + {b^{2/3}}} \right)^3}$. Since $f = {L^2}$, thus $L = {\left( {{h^{2/3}} + {b^{2/3}}} \right)^{3/2}}$. For $x,y,b,h > 0$, the function $f\left( {x,y} \right) = {\left( {x + b} \right)^2} + {\left( {y + h} \right)^2}$ is increasing. Hence, $L = {\left( {{h^{2/3}} + {b^{2/3}}} \right)^{3/2}}$ is a minimum length of the ladder subject to the constraint. (b) We have the square of the length function $f\left( {x,y} \right) = {\left( {x + b} \right)^2} + {\left( {y + h} \right)^2}$. In part (a) we have shown that the minimum value of $f$ is ${L^2} = {\left( {{h^{2/3}} + {b^{2/3}}} \right)^3}$. Thus, the level curve of $f$ at ${L^2}$ is given by ${\left( {x + b} \right)^2} + {\left( {y + h} \right)^2} = {\left( {{h^{2/3}} + {b^{2/3}}} \right)^3}$ Notice that this is the equation of a circle with radius $L = {\left( {{h^{2/3}} + {b^{2/3}}} \right)^{3/2}}$ centered at $\left( { - b, - h} \right)$. Since $\nabla f = \lambda \nabla g$, so the gradient $\nabla f$ is parallel to the gradient $\nabla g$. But $\nabla g$ is perpendicular to the tangent of the constraint $g\left( {x,y} \right) = xy - bh = 0$. So, $\nabla f$ is also perpendicular to the tangent of $g$. Thus, the level curve of $f$ at ${L^2}$ is tangent to the graph of $xy = bh$. This is illustrated in the figure attached. Hence, we conclude that the value of $L$ is also equal to the radius of the circle with center $\left( { - b, - h} \right)$ that is tangent to the graph of $xy = bh$.