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Names: / / /
Worksheet
Math 221
Instructions: Put the first and last name of everyone in your group at the top of your paper. Everyone is to do their own worksheet but only one from each group is graded with the score shared. Be sure to show your work and explain your reasoning.
Some questions on this worksheet reference a computer-generated demonstration (access it through Moodle) to aid in conceptual understanding of the questions.
- In this problem, determine the largest rectangle that fits inside the first quadrant and beneath the curve f (x) =
9 − x^2 = 3
(x 3
(a) Provide a sketch of f (x) as it appears in the first quadrant. Label your graph!
Answer:
(b) If one corner of an inscribed rectangle is at the point (0, 0) and the opposite corner is on the curve f (x), what is the area of the rectangle with base length x?
Answer: A = x ∗ f (x) (1) (c) Based on the demonstration, what is the approximate length of the base of the largest rectangle?
Answer: It is about x=2. (d) Use calculus to determine the exact dimensions of the largest inscribed rectan- gle. Compare a numerical approximation of your value to the approximation you found in the previous step.
Answer:
A = xf (x) = x(
9 − x^2 ) → A′(x) =
9 − x^2 + x
−x √ 9 − x^2
Setting this equation equal to zero and solving for the x-value, we find that the exact value is x =
9
- This is approximately 2.12.
(e) Building from your responses above, what are the dimensions of the largest rect- angle that may be inscribed inside a circle of radius 3?
Answer: The inscribed rectangle is largest when it is a square, as can be seen from the equation. If we plug in our x-value to the function, we get the same value back, so the dimensions of the square are 2
9 2 by 2
9
In this problem, determine the largest rectangle that fits inside the ellipse
(x 5
(y 3
1 by first considering the largest rectangle in the first quadrant beneath the curve f (x) =
9 − 9 x 2 25 = 3
(x 5
(a) Based on the demonstration, what is the approximate length of the base of the largest rectangle?
Answer: It looks like the largest rectangle has base 3.51. (b) Use calculus to determine the exact dimensions of the largest inscribed rectan- gle. Compare a numerical approximation of your value to the approximation you found in the previous step.
Answer:
A = xf (x) = x
x^2 25
→ A′(x) =
x^2 25
−^3 x 25
1 − x 2 25
Setting this equal to zero and solving for the x-value, we get a value of x =
25
This is approximately equal to 3.53. (c) Building from your responses above, what are the dimensions of the largest rect- angle that may be inscribed inside an ellipse with equation
(x 5
(y 3
Answer: By plugging in this value to the function, we get that the dimensions of the largest rectangle are 2
25 2 by 2
9
(d) Conjecture the dimensions of the largest rectangle that may be inscribed inside an ellipse with equation
(x a
(y b
Answer: By doing the same process, we can find the general form of the x-value. We do this by solving the general derivative.
A′(x) =
b^2 −
b^2 x^2 a^2
− x
xb
2
a^2
b^2 − b (^2) x 2 a^2
Setting this equal to zero and solving, we get the general base of the rectangle will be x =
a^2 2 =^ √a
- Therefore, the height will be^ f^ (x) =
b^2 2 =^ √b
linearly with the height. Therefore, the weighting of the variables is different in the two different cases.
- A 3f t tall TV is hung on a wall so that the lower edge is 4f t above eye level. How far from the wall should someone sit to get a maximum viewing angle θ? (Hint: as shown in the demonstration, θ = β − α. In order to maximize θ, you will first need to determine formulas for α and β that use the arctan function so you can relate θ back to x, the distance a person should sit from the wall.)
Answer: Define three angles. α is the angle from the viewer to the bottom of the tv and to the ground, β is the angle from the viewer to the top of the tv and to the ground, and θ is the angle from the viewer to top of the tv and to the bottom of the tv. Therefore, α + θ = β. We seek to maximize the viewing angle θ, so we rewrite this equation and find relations for the other two angles.
β − α = θ, tan β =
x
, tan α =
x
We now find the derivative in order to maximize the viewing angle with respect to the distance away from the TV.
dθ dx
d dx
arctan
x
− arctan
x
x^2 + 49
x^2 + 16
Setting this equation equal to zero and finding the x value, we get
7(x^2 + 16) = 4(x^2 + 49) ⇒ 3 x^2 = 4 · 49 − 7 · 16 = 4(49 − 28) = 4(21) = 84 ⇒ x^2 = 28.
- Determine the point on the line y = 3x − 5 that is closest to the point (− 7 , 15).
Answer: We seek to minimize the distance formula, but that will happen at the same time as the minimum of the distance squared.
z =
(x + 7)^2 + (3x − 20)^2 z^2 = (x + 7)^2 + (3x − 20)^2
Differentiating, we have
2 z
dz dx
= 2(x + 7) + 2(3x − 20) · 3 dz dx
x + 7 + 9x − 60 z
10 x − 53 z
Setting this equal to zero, we obtain x = 5310. We may use the First Derivative Test to verify that this is the x-coordinate for the closest point on the line.
