Turning a corner with a pole a. What is the length of the longest pole that can be carried horizontally around a corner at which a 3 - ft corridor and a 4-ft corridor meet at right angles? b. What is the length of the longest pole that can be carried horizontally around a corner at which a corridor that is ?a? ft wide and a corridor that is ?b? ft wide meet at right angles? c. What is the length of the longest pole that can be carried horizontally around a corner at which a corridor that is ?a? = 5 ft wide and a corridor that is ?b? = 5 ft wide meet at an angle of 120°? d. What is the length of the longest pole that can be carried around a corner at which a corridor that is ?a? ft wide and a corridor that is ?b? ft wide meet at right angles, assuming there is an 8-ft ceiling and that you may tilt the pole at any angle?

Solution 56E Step 1: (a)Consider a pole carried horizontally around a corner at which a 3-ft corridor and a 4-ft corridor meet at right angles. Consider the length of th e pole is l. So, at the corridor they meet at right angle. Consider the length is formed with one corridor is x and another corridor is y. So, you get Step 2 So, the length function by pythagoras theorem will be as l = +y 2 Now, the derivative form, dl d dx = dx( +y ) 2 1 d = 2 2(dx(x +y )) 2x +y dl 12x+2dx dx = 2 x +y2 Now, the corner of the floor is the constrained. So, you can say the constraints are The slope of the corner is dy dl 12x+2ydx Put this into dx = 2 x +y Now, the maximized condition is: So, y = 3x 4 Putx = 4 2 2 Put into the equationl = x y : So, the maximum length of the pole is5 ft (b)Consider a pole carried horizontally around a corner at which a a-ft corridor and a b corridor meet at right angles. Consider the length of the pole is l. So, at the corridor they meet at right angle. Consider the length is formed with one corridor is x and another corridor is y Step 3 So, the length function will be as: 2 2 l = +y Now, the derivative form, dl d dx = dx ( +y ) 2 = 12 2( dx(x +y )) 2x +y 2x+2ydy dx = 2 dx x +y Now, the corner of the floor is the constrained. So, you can say the constraints are: x = a The slope of the corner is: dl d 2 2 Put this into dx = dx( +y ) dl bxay dx = b x +y2 Now, the maximized condition is: So, bxay bx +y2 = 0 bxay = 0 y = bx a Put : Step 6 So, the length function will be as: Now, the derivative form, Now, the corner of the floor is the constrained. So, you can say the constraints are: The slope of the corner is: Put this into : Now, the maximized condition is: So, Put : Put into the equation : Now, this is actually ignoring the ceiling. Now, include the ceiling which is 8 ft tall. So, calculate in the similar way, then the length of the rod will be as: Put and calculate: So, the maximum length of the pole is: a +b +64 ft