Could you explain the second part please
Hi Edward — see below, where I’ve just right now answered that exact same question.
Hi, could you explain your answer to the second part of the question? Thanks!
And also, are these the minimum areas that you found? I.e if x was at a/2 as opposed to where it is on the diagram then that would be the max possible area for the 0≤x≤a/2 ?
Not sure I understand what you’re saying here? Can you explain?
First, we are expressing y in terms of x. When x goes between a/2 and a, we can see that y goes from (a√3)/2 to zero. So we get the equation y = (a-x)√3.
Second, we are saying that the area covered can be calculated by taking the area of the whole equilateral triangle (which is (a^2√3)/4 and subtracting the area of the triangle that we haven’t yet reached (which is (a-x)y/2).
We can use the first equation to substitute for y in the second equation, and then we just need to simplify the expression for A.
Thank you! This is really useful!
Hi, it states at the top that y= x * sqrt(3), I don’t really understand where this comes from, can you explain?
Hi Frances, thanks for your question. The triangle is equilateral, so when x <= a/2 the line is sloping up at π/3 radians (60 degrees). So the gradient of the line is tan π/3, which is equal to sqrt(3), so when x <= a/2, y = x * sqrt(3).
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