I've been trying to work out this question for hours and I can't seem to figure it out.
Question: "Point 'A' is situated in the bottom left hand corner of a vertical rectangle with the dimensions 6m x 3m. Point 'B' is situated in the opposite top corner of the rectangle. What is the minimum distance that could connect the two points of this right angled triangle? You must find this answer using optimisation and calculus."
I have worked out the answer is 6.708 metres using Pythagoras' Theorem, however I can't seem to get this answer using optimisation and calculus. I would assume you would begin with the following:
d = square root(x^2+y^2)
Where d= distance
Let's put the rectangle on the x-y grid
so that A is (0,0) and B is (3,6)
the equation of the line from A to B is clearly
y = 2x
d = √(x^2 + y^2)
= (x^2 + 4x^2)^(1/2) = √(5x^2) = (√5)x
dd/dx = √5
So for each unit of x, the length changes by √5
x goes from 0 to 3,
clearly when x = 0 , d = √5(0) = 0 , (the minimum ? )
when x = 3 , d = √5(3) = appr 6.708
Thank you so much for your help on this :)
Just a question though..
After you say y = 2x, you sub in y as 4x? ie. =(x^2+4x^2)^(1/2)
where did the '4x' come from?
since we had y^2
subbing in 2x for y would give us (2x)^2 = (2x)(2x) = 4x^2
notice it was 4x^2 and not 4x like you stated
Oh I see, thankyou for your help!!
To find the minimum distance between points A and B using optimization and calculus, you need to minimize the distance function d = √(x^2 + y^2). In this case, x represents the horizontal distance from point A to B, and y represents the vertical distance.
To start, you can establish some constraints. The rectangle has dimensions 6m x 3m, so you know that 0 ≤ x ≤ 6 and 0 ≤ y ≤ 3.
To find the minimum distance, you can set up an optimization problem. One approach is to use the method of Lagrange Multipliers. Firstly, introduce a Lagrange multiplier λ to account for the constraint equation of the rectangle, which is x*y - 18 = 0.
Formulate the Lagrangian function L(x, y, λ) as follows:
L(x, y, λ) = √(x^2 + y^2) + λ(x*y - 18)
To find the minimum distance, you will need to solve the partial derivatives of L(x, y, λ) with respect to x, y, and λ. Set all three derivatives equal to zero:
∂L/∂x = 0
∂L/∂y = 0
∂L/∂λ = 0
Once you solve these equations simultaneously, you will obtain values for x, y, and λ. Plug those values back into the distance function d = √(x^2 + y^2) to find the minimum distance.
However, it's worth noting that in this particular case, where point A is in the bottom-left corner (0, 0) and point B is in the top-right corner (6, 3), the minimum distance coincides with the length of the diagonal of the rectangle. The diagonal can be calculated using the Pythagorean theorem:
d = √(6^2 + 3^2) = √45 ≈ 6.708 meters
In this case, optimization and calculus are not necessary since the minimum distance is just the diagonal.