Mathematics, 04.11.2019 22:31, Riley3833

Awire 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 segment to the right of the cut is formed into a square. let x be the length of the segment to the left of the cut.

(a)-what value of x maximizes the total area of the two shapes?
(b)-what value of x minimizes the total area?

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answered: Jade1214

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

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 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.

Awire of length l is cut at the red point. the segment to the left of the cut is formed into an equi

answered: Guest

the range of a function is the range of y-values for which it is defined, or exists. given that we are only looking at the y-values, we can already discount the first two answers as they deal with the domain of the function (range of x-values).

now, what we need to do is find the smallest and highest y-values for which the function is defined. if you'd like, you could take a ruler and start at the very bottom of the graph, find the lowest point, then move the ruler higher and higher until you reach the highest point. doing so either in your head or with an actual ruler, we can see that the lowest y-value is -5 and the highest y-value is -1.

the question now remains as to whether they are included in the range (fourth answer) or excluded (third answer). the key here is to recognise whether they are marked with an open or a closed circle. since both (-2, -5) and (2, -1) are marked with a closed circle, they are points that exist on the graph and are thus included in the range (note that if they were marked with an open circle like (4, -2) then that would signify that there is a break in the graph at that point).

thus, the fourth answer, (y | -5 ≤ y ≤ -1), is correct.

also note that usually it is good practice to scan your eye inbetween the highest and lowest point for any breaks in the graph - we do have one here at (4, -2), however -2 is still included in the range because there is another point that is included in the graph with the same y-value, ie. point (1, -2).

i hope that but if you require any further explanation don't hesitate to comment below.