SOS...help!!! please email me at 969e221 at g mail. com here is the question:
Find the trapezoid of largest area that can be inscribed in the upper half of the ellipse x2/9 + y2/4 = 1 where the lower base of the trapezoid is on the x-axis and is 6 units long. I need to graph the ellipse and draw an inscribed trapezoid. Write and area formula for the trapezoid in x. Differentiate, find critical numbers, and apply an appropriate derivative test to show that you have found the trapezoid of largest area. HELP!!!!!!!!!!!!!!!!
well shucks. The base of the trapezoid is just the long axis of the ellipse, so you want to find two points (-x,y) and (x,y) on the ellipse which gives maximum area
a = 1/2(2x+6)y = (x+3)y
Now, we know that x^2/9 + y^2/4 = 1, so
y^2 = 4(1 - x^2/9)
so y = 2/3 √(9-x^2)
a = 2/3 (x+3)√(9-x^2)
da/dx = -(2x^2+3x-9)/√(9-x^2)
= -(x+3)(2x-3)/√(9-x^2)
This achieves a max at x=3/2.
We don't really need to find that a" < 0 at x=3/2, since with x = -3, a=0, a min.
Anyway, now just solve for y at x = 3/2 to find the area. The trapezoid has a top base of 3 and a bottom base of 6, with height = y.
omg thanks
I understand that you need help finding the trapezoid of largest area that can be inscribed in the upper half of the given ellipse and verifying it using calculus. I'll explain the step-by-step approach to solving this problem.
Step 1: Graph the ellipse:
Start by graphing the given ellipse x^2/9 + y^2/4 = 1. This ellipse has a center at the origin (0, 0), a semi-major axis of length 3, and a semi-minor axis of length 2. Draw the upper half of the ellipse.
Step 2: Draw the trapezoid:
For the trapezoid to be inscribed in the upper half of the ellipse, its base has to lie on the x-axis. Given that the length of the lower base is 6 units, draw a line segment on the x-axis with length 6.
Step 3: Find the coordinates of the vertices:
To find the other three vertices of the trapezoid, draw two straight lines from the endpoints of the lower base to intersect the ellipse in the upper half. The points where they intersect the ellipse will be the remaining vertices of the trapezoid.
Step 4: Write an area formula for the trapezoid in terms of x:
Let the trapezoid have vertices A(x1, 0), B(3cosθ, 2sinθ), C(-3cosθ, 2sinθ), and D(x2, 0), where θ is the angle between the positive x-axis and the point B. The lengths of the trapezoid's bases are AB = 6 and CD = x2 - x1.
The area (A) of a trapezoid is given by the formula: A = (1/2) * (sum of the parallel sides) * (height). In this case, the height of the trapezoid is given by the y-coordinate of points B or C, which is 2sinθ.
Therefore, the area of the trapezoid can be written as: A = (1/2) * (AB + CD) * 2sinθ = (AB + CD) * sinθ.
Step 5: Differentiate and find critical numbers:
Differentiate the area formula, A, with respect to x and find its critical numbers. To do this, replace sinθ with √(1 - (x^2 / 9)) in the area formula. Differentiate A with respect to x, set the derivative equal to zero, and solve for x to find the critical numbers.
Step 6: Apply the derivative test:
Evaluate the second derivative of A and determine its sign at the critical numbers. Use the second derivative test to determine whether the critical numbers correspond to a maximum or minimum.
Step 7: Find the trapezoid of largest area:
Evaluate the area of the trapezoid using the values of x found at the critical numbers. Compare the areas to determine which trapezoid has the largest area.
I hope this explanation helps you understand the process of solving the problem. If you need any further clarification or have additional questions, feel free to ask.