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yawa3891 [41]
3 years ago
6

Gina has 1200 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed ar

ea. What is the maximum​ area?
Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
8 0
We know the perimeter formula would be 1200=2x+2y. And the area will be A=x*y
We can to take the derivative of the area formula to find where it is a max but we must first substitute something in from the first formula so we only have one variable.

1200-2x=2y
600-x=y

A=(x)*(600-x)
A=600x-x^2

Now we take the derivative:
A'=600-2x (set equal to 0 and solve)
0=600-2x
2x=600
x=300 

Then when we plug this into the perimeter formula, we can solve for y
1200=2x+2y
1200=2(300)+2y
1200=600+2y
600=2y
y=300

So both the length and width are 300 yards, and the area would be 90,000 square yards

Hope this helps
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Can somebody tell me what the answer to y-13=8 is? Please and thank you.
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umka2103 [35]

Answer:

1 - (π/6) - (√3/4)

Step-by-step explanation:

First, calculate the area of ABCD. That's just 1*1 = 1 un^{2}

Then, we are told that arcs BD and AC are circular, meaning they are portions of a circle. Therefore, let's redraw this image a little. Look below for the image.

Essentially, what I've done is created an equilateral triangle and two sectors. You may be wondering why the triangle is equilateral. Well, we firstly know it's isosceles because the two original arcs are circular and congruent. Their overlap creates two congruent lines that can be extend from point M to the vertices A and D. These two lines also happen to be radii of the quarter circles because the two sectors have AM AND AD as radii. AM and AD = 1, so AMD is an equilateral triangle.

Draw altitude OD. This is a right angle. Since equilateral triangles are also isosceles triangles, and isosceles triangles have the property of sharing the median to base and altitude to base, we can say that OD = \frac{1}{2} AD = \frac{1}{2}. MD = 1, so this triangle is a 30-60-90 triangle. Why? This is because of the 30º angle converse theorem. The shorter leg is half of the hypotenuse MD, so <OMD is a 30º angle, making <ODM a 60º angle and inevitably <MDC a 30º angle (because <D is a right angle) and <ABM as well because the two sectors are congruent.

Because m<ABM = m<MDC = 30º, these two sectors are 30/360 of the individual circles or 1/12 of the original circles, which have areas of π (because 1^2*π = π). Each is 1/12π, combined they are 1/6π.

The area of the equilateral triangle is √3/8. This is because MO can be calculated, using the 30-60-90 ratios, to be √3/2. (1 * √3/2)/2 = (√3/2)/2 = √3/4; this uses the area of a triangle. Now, time to calculate the shaded area.

The area of the square is 1, the area of the sectors π/6, and the equilateral triangle √3/4. Subtract the equilateral triangle and sectors' areas from the area of the square to be left with the remaining shaded part. 1-(π/6)-(√3/4) is your final answer. To put that a little more cleanly below in the image.

Hope this helps, have a great day.

3 0
2 years ago
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