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ANEK [815]
3 years ago
6

What is the area of the largest rectangle that can be inscribed in an isosceles triangle with side lengths $8$, $\sqrt{80}$, and

$\sqrt{80}$? Assume that one side of the rectangle lies along the base of the triangle.

Mathematics
1 answer:
e-lub [12.9K]3 years ago
4 0
First let's solve the general case for an isosceles triangle of base 2 and height k. If the triangle is drawn so the base is on the x-axis and the apex is on the y-axis, the equation for the line containing the right side is
  y = -k(x-1)
Then a rectangle with x-dimension w will have an area that is the product of this width and the height y = -k((w/2)-1).
  area = w(-k(w/2 -1)) = (-k/2)w² +kw
The derivative of area with respect to w will be zero where the area is a maximum:
  d(area)/dw = 0 = -kw +k
  w = 1 . . . . . . add kw and divide by k
Using the formula for area, we find
  area = 1(-k(1/2-1)) = k/2
This value is 1/2 the total area of the triangle we started with.

If the side lengths of your triangle are 8, √80, and √80, the height will be given by the Pythagorean theorem as
  h = √((√80)² - (1/2·8)²) = √(80-16) = 8
Your triangle's area will be
  triangle area = (1/2)(8)(8) = 32

The maximum area of the inscribed rectangle will be half this value,
  16 square units

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Assume that f(x)=ln(1+x) is the given function and that Pn represents the nth Taylor Polynomial centered at x=0. Find the least
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Answer:

the least integer for n is 2

Step-by-step explanation:

We are given;

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centered at x=0

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We will use the format;

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2nd derivative: f"(x) = -1/(x + 1)² < 1! = 1

3rd derivative: f"'(x) = 2/(x + 1)³ < 2! = 2

4th derivative: f""(x) = -6/(x + 1)⁴ < 3! = 6

This follows that;

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Thus, error is;

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Answer:

Step-by-step explanation:

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