Help Week 10 F EXAMPLE 5 Find the area of the that can be EXAMPLE 5 Find the are

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Help Week 10 F EXAMPLE 5 Find the area of the that can be EXAMPLE 5 Find the area of the largest rectangle that can be inscribed in a semicircle of radius r SOLUTION 1 Let's take the semicircle to be the upper half of the circle xy- with center the vertices on the x-axis as shown in the top figure. Let (x, ) be the vertex that lies in the first quadrant. Then the rectangle has sides of lengths 2x and y, word inscribed means that the rectangle has two vertices on the semicircle and two So its area is A = raina Video Example To eliminate y we use the fact that (x. y) lies on the circle x y and so y . Thus Online Textbook A= The domain of this function is 0sxsr.Its derivative is which is 0 when 2x2 ,2, that is, x (since x 2 0). This value of x gives a maximum value of A since A(O) -0 and A(r) -0. Therefore the area A(- ) = 2 Let be the TION 2 A simpler solution is possible if we think of using an angle as a 8

Explanation / Answer

Let circle have equation: x² + y² = r²
Semicircle above x-axis has equation: y = (r² - x²) ----> 2nd blank

Let top right vertex of rectangle be in quadrant 1 at point (x, (r²-x²)), 0 < x < r
Rectangle has width 2x and height y = (r² - x²)

A = 2xy ---> 1st blank

A(x) = 2x (r² - x²) ---> 3 rd blank

A'(x) = 2(r² - x²) + 2x * 1/2 (r² - x²)^(-1/2) * -2x
A'(x) = 2(r² - x²) - 2x²/(r² - x²)

A'(x) = 0
2(r² - x²) - 2x²/(r² - x²) = 0
2(r² - x²) = 2x²/(r² - x²)
2(r² - x²) = 2x²
r² - x² = x²
2x² = r²
x² = r²/2
x = r/2 -------------> 4 th blank

A(x) = 2x (r² - x²)
A(r/2) = 2r/2 (r² - r²/2) -------> 5 th blank = (r² - r²/2)

= 2 r (r²/2) = 2 r * r/2 = r² -----------> 6 th blank = r²

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