an isosceles triangle has a vertex at the origin. Determine the area of the largest such triangle that is bound by the function x squared + 6y = 48 and explain why this is infact the maximum area.
because of the symmetry of the parabola, the base of the triangle must be parallel to the x-axis if the two sides are to be equal to form an isosceles triangle.
let the point of contact in the first quadrant be (x,y)
so the base of the triangle is 2x and its height is y
Area = 2xy
= 2x(8 - x^2/6)
= 16x - (1/3)x^3
d(Area)/dx = 16 - x^2= 0 for a max of area
x^2 = 16
x = ±4 ( and y = 16/3 )
so the largest area is
16(4) - 64/3 = 128/3 or 42 2/3
( I am assuming that the triangle lies above the x-axis. If you want the triangle to be located below the x-axis, of course the area would be without limits.
Make a sketch to understand this situation.
To find the area of the largest isosceles triangle bounded by the function x^2 + 6y = 48, we need to first determine the coordinates of its base.
Step 1: Rewrite the equation in terms of y:
x^2 + 6y = 48
6y = 48 - x^2
y = (48 - x^2) / 6
Step 2: Find the coordinates of the base points. Since we have an isosceles triangle with a vertex at the origin, the base will be symmetric with respect to the y-axis. So we just need to find the x-coordinate of one base point, and then we can find the y-coordinate using the equation obtained in Step 1.
Step 3: Determine the x-coordinate of one base point. To do this, we need to maximize the area of the triangle. Since the area of a triangle is 1/2 * base * height, we need to maximize the base length.
The base of the triangle is the distance between the two base points. Let's assume one of the base points has the coordinate (x, y).
The distance between the origin (0, 0) and the point (x, y) is given by the distance formula:
d = sqrt(x^2 + y^2).
To maximize the base length, we need to maximize the distance d. However, since y = (48 - x^2) / 6, we can substitute this value into the distance formula to express d in terms of x only.
d = sqrt(x^2 + [(48 - x^2) / 6]^2)
d = sqrt(x^2 + (48 - x^2)^2 / 36)
d = sqrt(x^2 + (2304 - 96x^2 + x^4) / 1296)
d = sqrt((x^4 - 96x^2 + x^2 + 2304) / 1296)
d = sqrt((x^4 - 95x^2 + 2304) / 1296)
Now, to find the maximum value of d, we need to find the critical points. We do this by finding the derivative of d with respect to x, and setting it equal to 0:
d' = (1/2) * (2x^3 - 190x) / (1296 * sqrt((x^4 - 95x^2 + 2304)/1296))
0 = (1/2) * (2x^3 - 190x) / (1296 * sqrt((x^4 - 95x^2 + 2304)/1296))
Simplifying the equation gives:
x^3 - 95x = 0
x(x^2 - 95) = 0
Therefore, the critical points are x = 0 and x = ±sqrt(95).
Since we are looking for the maximum value, we only need to consider x = sqrt(95).
Step 4: Find the y-coordinate of the base point. Substitute x = sqrt(95) into the equation y = (48 - x^2) / 6, obtained in Step 1:
y = (48 - (sqrt(95))^2) / 6
y = (48 - 95) / 6
y = -47 / 6
Thus, one base point is located at (sqrt(95), -47/6). Since the triangle is symmetric with respect to the y-axis, the other base point will be (-sqrt(95), -47/6).
Step 5: Find the height of the triangle. The height of an isosceles triangle is the perpendicular distance from the vertex to the base. In this case, the vertex is at the origin (0, 0), so the height is simply the y-coordinate of the base point, which is -47/6.
Step 6: Compute the area. The area of a triangle is given by the formula A = 1/2 * base * height. In this case, the base is the distance between the two base points, which is 2 * sqrt(95), and the height is -47/6.
A = 1/2 * 2 * sqrt(95) * (-47/6)
A = -47/6 * sqrt(95)
So, the area of the largest isosceles triangle bound by the function x^2 + 6y = 48 is (-47/6 * sqrt(95)).
To explain why this is indeed the maximum area, we can show that any other triangle that can be formed within the same boundary will have a smaller area.
We have maximized the base length by considering the critical points of the distance formula. Any other base length within the boundary will be smaller. Since the height of the triangle is fixed, any decrease in the base length will result in a smaller area.
Therefore, the triangle with the base points at (sqrt(95), -47/6) and (-sqrt(95), -47/6) has the largest area among all isosceles triangles bounded by the function x^2 + 6y = 48.
To determine the area of the largest isosceles triangle bound by the function x² + 6y = 48, we need to find the maximum area first and then explain why this is indeed the maximum area.
Step 1: Find the equation for the isosceles triangle
Since the vertex of the isosceles triangle is at the origin, let's assume that the other two vertices lie on the line y = mx. The reason for choosing this line is that all points on it will be equidistant from the origin, satisfying the condition of an isosceles triangle.
Step 2: Find the intersection points of the line and the given curve
Substitute y = mx into the equation x² + 6y = 48:
x² + 6mx = 48
Rearrange the equation:
x² + (6m)x - 48 = 0
Using the quadratic formula, we can find the x-values of the intersection points:
x = [-b ± √(b² - 4ac)] / 2a
Substituting a = 1, b = 6m, and c = -48 into the formula, we get:
x = [-6m ± √((6m)² - 4(-48))] / 2
Simplifying further:
x = [-6m ± √(36m² + 192)] / 2
x = -3m ± √(9m² + 48)
Step 3: Calculate the y-values of the intersection points
To find the y-values corresponding to the x-values obtained in Step 2, substitute the x-values into the equation y = mx:
y = m(-3m ± √(9m² + 48))
y = -3m² ± m√(9m² + 48)
Step 4: Calculate the base length of the triangle
As the triangle is isosceles, the base length will be twice the distance between the x-values of the intersection points.
Base length = 2 * |(-3m + √(9m² + 48)) - (-3m - √(9m² + 48))|
Base length = 2 * 2√(9m² + 48)
Base length = 4√(9m² + 48)
Step 5: Calculate the height of the triangle
The height of the triangle can be calculated by finding the distance between the y-intercept of the line (0,0) and the y-coordinate of one of the intersection points.
Height = |y-coordinate of one of the intersection points - 0|
Height = |(-3m + √(9m² + 48)) - 0|
Height = |-3m + √(9m² + 48)|
Height = 3m - √(9m² + 48)
Step 6: Calculate the area of the triangle
The area of the triangle is given by:
Area = (1/2) * base length * height
Area = (1/2) * 4√(9m² + 48) * (3m - √(9m² + 48))
Area = 2√(9m² + 48) * (3m - √(9m² + 48))
Area = 6m√(9m² + 48) - 2(9m² + 48)
Step 7: Determine the maximum area
To find the maximum area, we need to find the critical points of the area function by taking the derivative of the area equation with respect to m, setting it equal to zero, and solving for m.
d(Area) / dm = 0
6(√(9m² + 48) - 6m / √(9m² + 48)) = 0
√(9m² + 48) = 6m
Square both sides:
9m² + 48 = 36m²
27m² = 48
m² = 48 / 27
m² = 16 / 9
m = ±4 / 3
Since we are looking for the maximum area, we consider the positive value of m. So, m = 4 / 3.
Finally, substitute m = 4 / 3 back into the area equation to find the maximum area:
Area = 6 * (4/3) * √(9(4/3)² + 48) - 2(9(4/3)² + 48)
Simplifying the above expression will give us the maximum area of the isosceles triangle.