Where to plant the rose? The mathematics of gardening.

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.. :D

Ron.
 
Yes- a great little puzzle. You're reminding me of all the the A level and engineering mathematics i've forgotten since i left uni.... For instance looking at your solution it's obvious that the derivative would help as when the gradient is 0 it's going to be the max theta. That ability to make little connections like that completely needs refreshing! I'm actually tempted to get some engineering maths books out of the loft and see if i can still do it!
 
Hello Rich,

Glad you liked the problem,....why not grab your Calculus and Engineering Mathematics books out and have a look through them again.. :D

I'll keep the questions coming with the level of difficulty gradually increasing. So far from memory all bar one have come from the mathematics that is covered during the first semester of year one while studying Electrical Engineering at University.

Ron.
 
quattro wrote,...
So what was the answer?
Click to expand...

Hello Richard,

The distance from the 5 metre wall that the rose should be planted in order to maximise the sunlight that it receives is 5 - 2 X 5^0.5 metres.....that is subtract twice the square root of 5 from 5.

Ron.
 
So its 5 - (2*2.236) from the 5M wall

i.e. 0.528M from the high wall - doesn't seem right somehow :?

I don't know the maths you are using but to put it closer than 2' from the highest wall would leave it in shadow for longer, or so my gut feeling tells me. I could be wrong 8)

Richard
 
quattro wrote,..
So its 5 - (2*2.236) from the 5M wall

i.e. 0.528M from the high wall - doesn't seem right somehow

I don't know the maths you are using but to put it closer than 2' from the highest wall would leave it in shadow for longer, or so my gut feeling tells me. I could be wrong
Click to expand...

Hello Richard,

What the question requires you to find is a point 'p' along the 3 metre line that links the 5 and 2 metre walls such that the angle subtended by straight lines which link the top of each wall to point 'p' is as large as possible. If you were to place the rose signified by 'p' at any other point along the line, then the angle subtended by the lines will be reduced.

You can experiment by making a scale drawing and placing 'p' at different points and measure the angle obtained. Calculus is an extremely powerful too and even if intuitively an answer may not appear correct the fact that it can be proven mathematically means it is most definitely correct.

Ron.
 
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