Complete the explanation of why it is possible to draw more than two different rectangles with an area of 36 square units, but it is not possible to draw more than two rectangles with an area of 55 square units. the lengths of the sides of the rectangle are whole numbers.
(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
-- Each side has the same length as the side directly across from it.
-- So there are two sides that are both the width of the rectangle,
and another two sides that are both the length of the rectangle.
-- In this particular rectangle, when you add up all four sides,
the sum is 32 feet.
-- The width of the rectangle is 9 feet. So there are two sides that are
both 9 feet. When you add those up, you have 18 feet.
-- The REST of the 32 feet belongs to the other two sides.
That much is (32 - 18) = 14 feet. That's what the other two
sides add up to when you add them.
-- The other two sides are equal. So each of them is half of 14 = 7 feet .
This rectangle, or the description of it, is slightly weird.
The length of the rectangle is shorter than the width.
But that's OK. Two sides are 9 feet, and two sides are 7 feet, and
when you add them all up, you get 32 feet, so our solution is correct.
Let them call the sides anything they want to.
⇒ for a maximum area l=5, L=7
dividing both sides by 2
L + W = (perimeter / 2)
Length = (perimeter / 2) -Width
Width = (perimeter / 2) -Length
The rectangle is 10 cm by 18 cm.
Let x represent the distance from the center of the rectangle to the longer side. Then the distance to the shorter side is x+4. The short side of the rectangle will have length 2x, and the long side will have length 2(x+4) = 2x+8.
The perimeter is twice the sum of the side lengths, so is ...
56 = 2(2x +(2x+8)) = 8x +16
40 = 8x . . . . . . . subtract 16
40/8 = x = 5 . . . divide by the coefficient of x
The side lengths are 2x=10 and 2x+8 = 18 centimeters.
Figure a. E_net = 99.518 N/C
Figure b. E_net = 177.151 N / C
- Attachment for figures missing in the question.
- The dimensions for rectangle are = 7.79 x 3.99 cm
- All four charges have equal magnitude Q = 10.6*10^-12 C
Find the magnitude of the electric field at the center of the rectangle in Figures a and b.
- The Electric field generated by an charged particle Q at a distance r is given by:
E = k*Q / r^2
- Where, k is the coulomb's constant = 8.99 * 10^9
- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:
E_1 + E_3 = 0
- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:
E_net = E_2 + E_4
E_2 = E_4
E_net = 2*E = 2*k*Q / r^2
- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:
r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )
r = sqrt ( 1.9151*10^-3 ) = 0.043762 m
- Plug the values in the E_net expression developed above:
E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3
E_net = 99.518 N/C
- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).
- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.
E_net = 2*E_part(a)*cos(Q)
- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:
Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.
- Now, compute the net electric field E_net:
E_net = 2*(99.518)*cos(27.12)
E_net = 177.151 N / C
Part a) Exactly values
The width of the rectangle is
The length of the rectangle is
Part b) Approximately values
The width of the rectangle is
The length of the rectangle is
see the attached figure to better understand the problem
we know that
The area of rectangle is equal to
substitute the values in the formula
Solve for x
Divide by 4 both sides
square root both sides
The solution is the positive value
The lengths of the sides are 20 cm and 20 cm
Perimeter, P = 80cm
Represent the length and width with L and W, respectively;
Substitute 80 for P
Divide through by 2
Make L the subject of formula
Area of a rectangle is calculated as thus;
Substitute 40 - B for L
Express this as a function
Set A(B) = 0 to determine the roots
The maximum area of a rectangle occurs at half the sum of the roots;
Hence the lengths of the sides are 20 cm and 20 cm
P = 2 * (a+b)
This means that sum of two sides is:
64 = 2* (a+b)
Pairs of numbers that add to 32 are:
We do not consider last pair as it would represent square. Areas these sides cover are:
Maximum area is 255 and lengths of sides are 15 and 17. This answer makes sense because larger sides give larger area.