A rectangle has its base on the x axis and its upper two vertices on the parabola y = 9 − x2. (a) Draw a graph of this problem.
(b) Label the upper right vertex of the rectangle (x,y) and indicate the lengths of the sides of the rectangle. (c) Express the area A as a function of x and state the domain of A. (d) Calculate the derviative of A and solve for the critical values. (e) What is the largest area the rectangle can have?
1 answer:
Answer:
- (a) see attached
- (b) width: 2x; height: y = 9-x²
- (c) A=2x(9-x²) . . . 0 ≤ x ≤ 3
- (d) dA/dx = -6x² +18; x=±√3
- (e) 12√3 units²
Step-by-step explanation:
(a) The attachment shows the graph of the parabola in blue. It also shows an inscribed rectangle in black.
(b) The upper right point of the rectangle is shown in the attachment as (x, y). The dimension y is the height of the rectangle. The x-dimension is half the width of the rectangle, which is symmetrical about the y-axis. Hence the width is 2x.
(c) As with any rectangle, the area is the product of length and width:
... A = (2x)(9 -x²) . . . . . the attachment shows a graph of this
... A = -2x³ +18x . . . . . expanded form suitable for differentiation
A suitable domain for A is where both x and A are non-negative: 0 ≤ x ≤ 3.
(d) The derivative of A with respect to x is ...
... A' = -6x² +18
This is defined everywhere, so the critical values will be where A' = 0.
... 0 = -6x² +18
... 3 = x² . . . . . . . divide by -6, add 3
... √3 = x . . . . . . . -√3 is also a solution, but is not in the domain of A
(e) The rectangle will have its largest area where x=√3. That area is ...
... A = 2x(9 -x²) = 2√3(9 -(√3)²) = 2√3(6)
... A = 12√3 . . . . square units . . . . ≈ 20.785 units²
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Answer:
Step-by-step explanation:
Given
Required
Fill in the box
From the question, the range is:
Range is calculated as:
From the box, we have:
So:
The box, becomes:
From the question:
--- interquartile range
This is calculated as:
is the median of the upper half while
is the median of the lower half.
So, we need to split the given boxes into two equal halves (7 each)
<u>Lower half:</u>
<u>Upper half</u>
<u></u><u></u>
The quartile is calculated by calculating the median for each of the above halves is calculated as:
Where N = 7
So, we have:
So,
= 4th item of the upper halves
= 4th item of the lower halves
From the upper halves
<u></u><u></u>
<u></u>
We have:
can not be determined from the lower halves because the 4th item is missing.
So, we make use of:
Where and
So:
So, the lower half becomes:
<u>Lower half:</u>
From this, the updated values of the box is:
From the question, the median is:
and
To calculate the median, we make use of:
This means that, the median is the average of the 7th and 8th items.
The 7th and 8th items are blanks.
However, from the question; the mode is:
Since the values of the box are in increasing order and the average of 18 and 18 do not equal 22 (i.e. the median), then the 7th item is:
The 8th item is calculated as thus:
Multiply through by 2
The updated values of the box is:
From the question.
Mean is calculated as:
So, we have:
Collect like terms
Multiply through by 14
This gives:
From the updated box,
We know that:
<em>The 2nd value can only be either 2 or 3</em>
<em>The 12th value can take any of the range 33 to 57</em>
Of these values, the only possible values of 2nd and 12th that give a sum of 60 are:
i.e.
So, the complete box is: