Hello Rich, Richard and Darth,
Ok with the angle subtended by the lines intersecting at 'p', given as theta = PI - arccot(p/5) - arccot((3 - p)/2), now find dtheta/dp as the position of 'p' is dependent upon the angle theta. The derivative yields 5/(25 + p^2) - 2/(4 + (3-p)^2). To find the critical point equate dtheta/dp to zero and solve for 'p'. A quadratic equation will ensue with solutions of 5 +/- 2 X 5^0.5. Use the first derivative test to test the solution along with the boundaries of 'p' within the closed interval [0,3]. Only 5 - 2 X 5^0.5 is a possible solution, so this gives the distance from the 5 m wall where the rose should be planted, thus maximising the angle theta.
Glad you liked the problem..
Ron.
Ok with the angle subtended by the lines intersecting at 'p', given as theta = PI - arccot(p/5) - arccot((3 - p)/2), now find dtheta/dp as the position of 'p' is dependent upon the angle theta. The derivative yields 5/(25 + p^2) - 2/(4 + (3-p)^2). To find the critical point equate dtheta/dp to zero and solve for 'p'. A quadratic equation will ensue with solutions of 5 +/- 2 X 5^0.5. Use the first derivative test to test the solution along with the boundaries of 'p' within the closed interval [0,3]. Only 5 - 2 X 5^0.5 is a possible solution, so this gives the distance from the 5 m wall where the rose should be planted, thus maximising the angle theta.
Glad you liked the problem..
Ron.