find the dimensions and the area of the largest rectangle that can be inscribed in the upper half of the…

find the dimensions and the area of the largest rectangle that can be inscribed in the upper half of the ellipse\n\\( \\frac { x ^ { 2 } } { a ^ { 2 } } + \\frac { y ^ { 2 } } { b ^ { 2 } } = 1 \\)\nnumerical exploration\nfor the default values of \\( a = 3 \\) and \\( b = 2.5 \\), what approximate value of \\( x \\) gives the largest rectangle area? (round your answer to two decimal places.)\nwhat are the dimensions of this rectangle? (enter the dimensions as a comma-separated list. round your answers to two decimal places.)\nwith \\( a = 3, b = 4 \\). what approximate value of \\( x \\) gives the largest rectangle area? (round your answer to two decimal places.)\nof this rectangl? (enter the dimensions as a comma-separated list. round your answers to two decimal places.)\nif you change the value of \\( b \\), does the same value of \\( x \\) give the rectangle with the largest area?\n\\( \\bigcirc \\) yes\n\\( \\bigcirc \\) no\nif you change the value of \\( a \\), does the same value of \\( x \\) give the rectangle with the largest area?\n\\( \\bigcirc \\) yes\n\\( \\bigcirc \\) no\ndoes the optimal rectangle depend on \\( a, b \\), or both?\n\\( \\bigcirc \\) the optimal rectangle depends on \\( a \\) only.\n\\( \\bigcirc \\) the optimal rectangle depends on \\( b \\) only.\n\\( \\bigcirc \\) the optimal rectangle depends on both \\( a \\) and \\( b \\).

find the dimensions and the area of the largest rectangle that can be inscribed in the upper half of the ellipse\n\\( \\frac { x ^ { 2 } } { a ^ { 2 } } + \\frac { y ^ { 2 } } { b ^ { 2 } } = 1 \\)\nnumerical exploration\nfor the default values of \\( a = 3 \\) and \\( b = 2.5 \\), what approximate value of \\( x \\) gives the largest rectangle area? (round your answer to two decimal places.)\nwhat are the dimensions of this rectangle? (enter the dimensions as a comma-separated list. round your answers to two decimal places.)\nwith \\( a = 3, b = 4 \\). what approximate value of \\( x \\) gives the largest rectangle area? (round your answer to two decimal places.)\nof this rectangl? (enter the dimensions as a comma-separated list. round your answers to two decimal places.)\nif you change the value of \\( b \\), does the same value of \\( x \\) give the rectangle with the largest area?\n\\( \\bigcirc \\) yes\n\\( \\bigcirc \\) no\nif you change the value of \\( a \\), does the same value of \\( x \\) give the rectangle with the largest area?\n\\( \\bigcirc \\) yes\n\\( \\bigcirc \\) no\ndoes the optimal rectangle depend on \\( a, b \\), or both?\n\\( \\bigcirc \\) the optimal rectangle depends on \\( a \\) only.\n\\( \\bigcirc \\) the optimal rectangle depends on \\( b \\) only.\n\\( \\bigcirc \\) the optimal rectangle depends on both \\( a \\) and \\( b \\).

Answer

Explanation:

Step1: Find the equation for the area of the rectangle

From the ellipse equation (\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1), we can express (y=b\sqrt{1-\frac{x^{2}}{a^{2}}}). The area of the rectangle (A = 2xy=2xb\sqrt{1-\frac{x^{2}}{a^{2}}}). For (a = 3) and (b=2.5), (A=5x\sqrt{1-\frac{x^{2}}{9}}).

Step2: Take the derivative of the area function

Using the product - rule ((uv)^\prime=u^\prime v + uv^\prime), let (u = 5x) and (v=\sqrt{1-\frac{x^{2}}{9}}=(1-\frac{x^{2}}{9})^{\frac{1}{2}}). Then (u^\prime=5) and (v^\prime=\frac{1}{2}(1-\frac{x^{2}}{9})^{-\frac{1}{2}}(-\frac{2x}{9})). (A^\prime=5\sqrt{1 - \frac{x^{2}}{9}}+5x\times\frac{1}{2}(1-\frac{x^{2}}{9})^{-\frac{1}{2}}(-\frac{2x}{9})) Simplify (A^\prime=\frac{45 - 10x^{2}}{3\sqrt{9 - x^{2}}}) Set (A^\prime = 0), then (45-10x^{2}=0), (x^{2}=\frac{45}{10}=4.5), (x=\sqrt{4.5}\approx2.12)

Step3: Find the dimensions of the rectangle

(y=b\sqrt{1-\frac{x^{2}}{a^{2}}}), when (a = 3), (b = 2.5) and (x\approx2.12) (y = 2.5\sqrt{1-\frac{(2.12)^{2}}{ 9}}\approx2.5\sqrt{1 - 0.5}\approx1 77) The length (l = 2x\approx4.24 ), the width (w=y\approx1.7 7) For (a = 3), (b = 4) \ (A = 8x\sqrt{1-\frac{x^{2}}{9}}) (A^\prime=\frac{7 2-16 x^{2}}{3\sqrt{9 - x^{2}}}\ ), set (A^\prime = 0), (72-16x^{2}=0), (x ^{2}=\frac{7 2}{16}=4.5), (x=\sqrt{4.5}\approx2.12) (y = 4\sqrt{1-\frac{(2.12)^{2}}{9}}\approx4\sqrt{ 0.5}\approx2.83) The length (l = 2x\approx4.24), the width (w=y\approx2.83) Since (x=\frac{a}{\sqrt{2}}), changing (b) does not change (x) (Yes for the first - change - (b) question). Changing (a) changes (x) (No for the change - (a) question). The optimal rectangle depends on (a) ( 2x is related to (a)) and (b) ( (y) is related to (b)), so it depends on both.

Answer: 1. (x\approx2.12)

  1. (4.24, 1.77)
  2. (x\approx2 12)
  3. (4. 24,2.83)
  4. Yes
  5. No
  6. The optimal rectangle depends on both (a) and (b)