the ratio of large triangles to small triangles is 2-6 .If there are 100 triangles in the design ,how many small triangles are t
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Answer:
Part a)
Part b) The side length x that give the maximum area is 120 meters
Part c) The maximum area is 14,400 square meters
Step-by-step explanation:
The picture of the question in the attached figure
Part a) Find a function that gives the area A(x) of the playground (in square meters) in terms of x
we know that
The perimeter of the rectangular playground is given by
we have
substitute
solve for W
<u><em>Find the area of the rectangular playground</em></u>
The area is given by
we have
substitute
Convert to function notation
Part b) What side length x gives the maximum area that the playground can have?
we have
This function represent a vertical parabola open downward (the leading coefficient is negative)
The vertex represent a maximum
The x-coordinate of the vertex represent the length that give the maximum area that the playground can have
Convert the quadratic equation into vertex form
Factor -1
Complete the square
The vertex is the point (120,14,400)
therefore
The side length x that give the maximum area is 120 meters
Part c) What is the maximum area that the playground can have?
we know that
The y-coordinate of the vertex represent the maximum area
The vertex is the point (120,14,400) -----> see part b)
therefore
The maximum area is 14,400 square meters
Verify
The playground is a square
The answers are shown in the attached image
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Explanation:
Set the denominator x^4-8x^3+16x^2 equal to zero and solve for x
x^4-8x^3+16x^2 = 0
x^2(x^2-8x+16) = 0
x^2(x-4)^2 = 0
x^2 = 0 or (x-4)^2 = 0
x = 0 or x-4 = 0
x = 0 or x = 4
The x values 0 and 4 make the denominator zero
These x values lead to asymptote discontinuities because the numerator 8x-24 = 8(x-3) has no common factors which cancel with the denominator factors.
There are two vertical asymptotes
Let's see what happens when we plug in a value to the left of x = 0, say x = -1, we'd get
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(-1) = (8(-1)-24)/((-1)^4-8(-1)^3+16(-1)^2)
f(-1) = -1.28
So as x gets closer and closer to x = 0 from the left side, the f(x) is heading to negative infinity
Now plug in some value to the right of x = 0. I'm going to pick x = 1
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(1) = (8(1)-24)/((1)^4-8(1)^3+16(1)^2)
f(1) = -1.78 (approximate)
So as x gets closer and closer to x = 0 from the right side, the f(x) is heading to negative infinity
Overall, as x approaches 0 from either the left or right side of x = 0, the y value is heading off to negative infinity
---------------------
Repeat for values to the left and right of x = 4
We can't use x = 1 as it turns out that x = 3 is a root
But we can use something like x = 3.5 to find that...
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(3.5) = (8(3.5)-24)/((3.5)^4-8(3.5)^3+16(3.5)^2)
f(3.5) = 1.31 approx
So as x gets closer to x = 4 from the left, y is getting closer to positive infinity
Plug in x = 5 to find that
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(5) = (8(5)-24)/((5)^4-8(5)^3+16(5)^2)
f(5) = 0.64
which has the same behavior as the left side
So overall, as we approach x = 4, the y value is heading off to positive infinity
Again everything is summarized in the image attachment
Note: you could make a table of more values but they would effectively say what has already been said. It would be redundant busy work. However, its always good practice for function evaluation.
-------------------------------------------------------------------------
Explanation:
Set the denominator x^4-8x^3+16x^2 equal to zero and solve for x
x^4-8x^3+16x^2 = 0
x^2(x^2-8x+16) = 0
x^2(x-4)^2 = 0
x^2 = 0 or (x-4)^2 = 0
x = 0 or x-4 = 0
x = 0 or x = 4
The x values 0 and 4 make the denominator zero
These x values lead to asymptote discontinuities because the numerator 8x-24 = 8(x-3) has no common factors which cancel with the denominator factors.
There are two vertical asymptotes
Let's see what happens when we plug in a value to the left of x = 0, say x = -1, we'd get
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(-1) = (8(-1)-24)/((-1)^4-8(-1)^3+16(-1)^2)
f(-1) = -1.28
So as x gets closer and closer to x = 0 from the left side, the f(x) is heading to negative infinity
Now plug in some value to the right of x = 0. I'm going to pick x = 1
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(1) = (8(1)-24)/((1)^4-8(1)^3+16(1)^2)
f(1) = -1.78 (approximate)
So as x gets closer and closer to x = 0 from the right side, the f(x) is heading to negative infinity
Overall, as x approaches 0 from either the left or right side of x = 0, the y value is heading off to negative infinity
---------------------
Repeat for values to the left and right of x = 4
We can't use x = 1 as it turns out that x = 3 is a root
But we can use something like x = 3.5 to find that...
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(3.5) = (8(3.5)-24)/((3.5)^4-8(3.5)^3+16(3.5)^2)
f(3.5) = 1.31 approx
So as x gets closer to x = 4 from the left, y is getting closer to positive infinity
Plug in x = 5 to find that
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(5) = (8(5)-24)/((5)^4-8(5)^3+16(5)^2)
f(5) = 0.64
which has the same behavior as the left side
So overall, as we approach x = 4, the y value is heading off to positive infinity
Again everything is summarized in the image attachment
Note: you could make a table of more values but they would effectively say what has already been said. It would be redundant busy work. However, its always good practice for function evaluation.