Two linear are parallel if and only if their slopes are

1 answer:

8 0

Their slopes must be the same, but with different y-intercepts.

Here is an example: 2x + 1 and 2x + 4 are parallel lines.

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A wire of length L is cut at the red point. The segment to the left of the cut is formed into an equilateral triangle, and the s

Elden [556K]

Answer:

$(a) x=0\\(b)\ x=\dfrac{4L}{4+\sqrt{3}}$

Step-by-step explanation:

Imagine that the wire is cut as shown in the picture. We know that the lenght of the segment to the left is $x$ and therefore the lenght of the segment to the right must be $L-x.$

Using the formula for the area of an equilateral triangle of sides of length x, we get that $A_{\bigtriangleup}(x)=\dfrac{\sqrt{3}}{4}\, x^2.$

And using the formula for the area of a square of length L-x, we obtain that $A_{\square}(x)=(L-x)^2.$

Then, the total area of the two shapes is giving by the sum of both areas: $A_T(x)=A_{\bigtriangleup}(x)+A_{\square}(x)=\dfrac{\sqrt{3}}{4}\, x^2 + (L-x)^2.$

Now we have to find the values x where the function $A_T(x)$ attains its maximum and its minimum. For this purpose, we calculate its critical points, which occurs when the derivative vanishes:

$A_T'(x)=A_{\bigtriangleup}'(x)+A_{\square}'(x)=\dfrac{\sqrt{3}}{2}x-2(L-x)=0.$

Solving for x we get that:

$x=\dfrac{4L}{4+\sqrt{3}}.$

This is the only critical point. Using the second derivative test we found that, since $A_T''(x)=\dfrac{1}{2} \left(4 + \sqrt{3}\right) >0,$ then at $x=\dfrac{4L}{4+\sqrt{3}}$ the area of the two shapes is minimized.

Now, the only way we have maximum area is when the red point is on the extreme of the wire, at x = 0. This because in that situation, there is no triangle that can be formed and therefore the area is equals $L^2.$