What is the maximum possible area of a rectangle with a base that lies on the X-axis and with two upper vertices that lie on the graph of the equation?

2 Answers. The maximum area is approximately 8.61 square units.

How do you find the maximum area of a rectangle in a semi circle?

Let r be the radius of the semicircle, x one half of the base of the rectangle, and y the height of the rectangle. We want to maximize the area, A = 2xy. Thus, the base of the rectangle has length = r/√2 and its height has length √2*r/2.

What is the area of the largest rectangle that can be inscribed in a semi circle of radius 10cm?

54 square inches
A rectangle is inscribed in a semicircle of radius 10 cm. What is the area of the largest rectangle we can inscribe? Amax = xw = (5 / 2)(10 / 2) = 100 Page 7 A poster is supposed to have margins of 1 inch on the left and right and 1.5 inches on top and on bottom. The printed area is supposed to be 54 square inches.

What is maximum area of a rectangle?

Approach: For area to be maximum of any rectangle the difference of length and breadth must be minimal. So, in such case the length must be ceil (perimeter / 4) and breadth will be be floor(perimeter /4). Hence the maximum area of a rectangle with given perimeter is equal to ceil(perimeter/4) * floor(perimeter/4).

What is the maximum area in square units of a rectangle?

32 square units
Hence, maximum area of rectangle is 32 square units.

What is the largest area of a rectangle inscribed in an ellipse?

Thus the maximum area of a rectangle that can be inscribed in an ellipse is 2ab sq. units.

What is the maximum area of a rectangle with diagonals?

This has a maximum when the diagonals are orthogonal, i.e. when the rectangle is a square, with side 2 r. Hence, the dimensions of the original rectangle are 2 r and r / 2. Thus, length of rectangle is x − ( − x) = 2 x and height is y. Area A = 2 x y. Maximizing A is equivalent to maximizing A 2 Therefore, max.

What is the maximum area of a rectangle inscribed in ellipse?

6 Answers. The vertices of any rectangle inscribed in an ellipse is given by (±acos(θ), ±bsin(θ)) The area of the rectangle is given by A(θ) = 4abcos(θ)sin(θ) = 2absin(2θ) Hence, the maximum is when sin(2θ) = 1. Hence, the maximum area is when 2θ = π 2 i.e. θ = π 4. The maximum area is A = 2ab.

How do you find the area of a rectangle and triangle?

So the area of the rectangle is 1/4 the base of the triangle times its height, so the rectangle is half the area of the triangle. Call the long side of the triangle the base of the triangle. Connect the midpoints of the opposite sides. That line is parallel to the base and forms one side of the rectangle with the base on the other side.

Can a rectangle with a side on AB be larger?

FG is half the length of AB, and FH, a side of the rectangle, is half the altitude CD of the triangle. So the area of the rectangle is 1/4 the base of the triangle times its height, so the rectangle is half the area of the triangle. No other rectangle with a side on AB can be larger.