Question

39. Let L be the minimum length of a ladder that can reach over a fence of height h to a wall located a distance b behind the wall. a. Use Lagrange multipliers to show that L = (h2/3 + 62/3;3/2 (Eigure 20). Hint: Show that the problem amounts to minimizing f (x, y) = (x + b)² + (y+h)? subject to y/b = h/x or ry = bh. b. Show that the value of L is also equal to the radius of the circle with center (-6, – h) that is tangent to the graph of XY = bh. -Xy=bh L L Wall h Ladder (-6, -h) b x Fence Rogawski et al., Multivariable Calculus, 4e, 2019 W. H. Freeman and Company FIGURE 20

Answer 1

Solution - (9) -by phthagoras- L The long thay the ladder can be written as- V (x+b)? Hy th}? We have to minimize L minimiziry L is same as minimizing (x +b)2 +8th)? 2 f(x,y) = 2² -bh co apply lagangis Now We howe for two similen tromgeles we com unte 9/ hla g (x, y) = xy bath f and So we multipliers and fx = ngx 2 (x+b) = my 0 fy = egy 2 lythl=ria from eqn 0 and 276 =yth a

constraint in the ear x²+bx = y2 + yh the value pulting of y equation of x2 +0x = (t ) 2 + ( a ) A 24 + bx²_ bh-b²h²=0 the equation we get- solving x= Ý khz putting the value x in constant equation g y= 3 b2h Length of ladles can be written as (69844 20 342 L = )