3) Consider rectangles located as shown in the first quadrant and inscribed under a decreasing curve, with the lower left hand corner at the origin and the upper right hand corner on the curve
y = sqrt(9 - x2)
Find the width, height and area of the largest such rectangle.
let the top right vertex of the rectangle which is on the curve by (x,y)
x is the width of the rectangle, y is the height
then the area = xy
=x√(9-x^2)
differentiate using the product rule, set the derivative equal to zero and solve for x.
or....
we could use common sense, notice that your given equation when squared and rearranged is really part of a circle with centre at the origin and radius 3.
So the largest rectangle would be where the x equals the y, namely when x = 3/√2
(which is what you get if you do it by Calculus as I decribed above.
So the area = (3/√2)^2 = 9/2
To find the width, height, and area of the largest rectangle inscribed under the curve y = sqrt(9 - x^2), you can use calculus or common sense.
The calculus approach involves differentiating the area equation, setting the derivative equal to zero, and solving for x.
1. Start with the area equation: A = x * sqrt(9 - x^2).
2. Differentiate the area equation with respect to x using the product rule:
dA/dx = (sqrt(9 - x^2)) + x * (1/2) * (9 - x^2)^(-1/2) * (-2x).
3. Set the derivative equal to zero and solve for x:
(sqrt(9 - x^2)) + x * (1/2) * (9 - x^2)^(-1/2) * (-2x) = 0.
4. Simplify and solve the equation for x. This will give you the value of x that maximizes the area.
5. Substitute the value of x into the original area equation to find the maximum area.
The common sense approach involves recognizing that the given equation, y = sqrt(9 - x^2), when squared and rearranged, represents part of a circle with center at the origin and radius 3. The largest rectangle will have its width equal to its height.
1. Observe that the maximum area occurs when x = y.
2. Solve for x by setting y = x:
sqrt(9 - x^2) = x.
3. Square both sides to get rid of the square root:
9 - x^2 = x^2.
4. Rearrange the equation to solve for x:
2x^2 = 9.
x^2 = 9/2.
x = √(9/2) = 3/√2.
5. Substitute the value of x into the area equation to find the maximum area:
A = (3/√2) * sqrt(9 - (3/√2)^2).
A = (3/√2) * sqrt(9 - 9/2).
A = (3/√2) * sqrt(9/2).
A = (3/√2) * (3/√2) * √2.
A = 9/2.
Therefore, the width, height, and area of the largest rectangle are x = 3/√2, y = 3/√2, and area = 9/2, respectively.