you can draw a square inside another square by placing each vertex of the inner square on one side of the outer square. The large square in this diagram has side length 10 units
a) Determine the area of the inscribed square as a function of x ( X is from one corner of the large square to the vertex of the inner square - the shorter section of the side which has been divided by two by the inner square)
b)range for area function
* I know the answer for a is A(x) = 2x² - 20x + 100 ---Answer
and I know how to do it.
But how do you find b. Some one said that the value of X can only be from 0-5. Why is that?
b) let's look at
A(x) = 2x^2 - 20x + 100
This is an upwards parabola.
I will assume you know how to find its vertex.
In standard form you would get
A(x) = 2(x-5)^2 + 50
with vertex at (5,50)
This means your smallest inner square has an area of 50 square units when x = 5, clearly at the midpoint of the original square.
Look back at your diagram,
it is obvious that the domain of our A(x) is
0 < x < 10
at x = 0, A(0) = 100
(the inner square coincides with the outer square)
at x = 5 , A(5) = 50 , our vertex (minimum)
at x = 10, A(10) = 100, back to same inner as outer
since range would be all the A(x)'s possible
the range is
50 < A(x) < 100
Ah, the joy of geometry! Let me explain the answer to part b) of your question.
To find the range for the area function, let's understand the limitations imposed by the problem.
When the vertex of the inner square coincides with a corner of the large square (i.e., x = 0), you'll notice that the inner square's side length is also 0. This means that the area of the inscribed square is 0 as well.
Now, as x increases, the side length of the inner square also increases. However, when x reaches half the length of the side of the large square (i.e., x = 5), the entire side of the inner square coincides with the side of the large square. At this point, the area of the inscribed square becomes maximum.
Think about it: Any larger value of x would cause the inscribed square to extend outside the boundaries of the large square. That wouldn't be aesthetically pleasing, would it? So, x should not exceed half the length of the side of the big square.
Therefore, the range for the area function would be from 0 to the maximum area attained when x = 5.
I hope this clarifies your doubt! And remember, as a wise clown once said, "Geometry may seem pointless, but it has its angles!"
To determine the range for the area function of the inscribed square, we need to consider the possible values for x. In this case, x represents the distance from one corner of the large square to the vertex of the inner square.
If x = 0, it means that the vertex of the inner square coincides with one of the corners of the large square. In this case, the inner square would have no area, since it would be degenerate and collapse into a single point.
If x = 5, it means that the vertex of the inner square lies exactly halfway along one side of the large square. In this case, the inner square would be at its maximum size, with a side length of 10 units, equal to the side length of the large square. Therefore, the maximum area of the inscribed square would be 10² = 100 square units.
However, it's important to note that as x increases beyond 5, the vertex of the inner square will no longer be on the side of the large square, but will instead cross over and lie within the interior of the large square. This violates the original condition that the inscribed square has each vertex on one side of the outer square. Therefore, any values of x greater than 5 are not valid in this context.
In conclusion, the valid range for x is 0 ≤ x ≤ 5, which corresponds to an area range of 0 square units to 100 square units.
To find the range for the area function, let's consider the properties of the square.
We know that any side length 'x' of the inner square can be no larger than half the length of the side of the outer square. In this case, the side length of the outer square is 10 units. Therefore, the maximum value for 'x' would be 5 units.
To understand why this is the case, let's visualize it. If we draw a diagram of the square with an inscribed square, with 'x' being the distance from one corner of the large square to the vertex of the inner square, we can see that 'x' can never be greater than half the length of the side of the outer square. If it were larger than that, the inner square would extend beyond the boundaries of the outer square.
So, the maximum value for 'x' is 5 units. However, there is no restriction on the lower limit for 'x'. It can be any positive value greater than 0.
Therefore, the range for the area function A(x) = 2x² - 20x + 100 is (0, 5]. This means that the area of the inscribed square can take any positive value greater than 0 and up to 5, inclusive.