A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y= 6-x^2. What are the dimensions of such a rectangle with the greatest possible area?
x^2 - 6 = - y
vertex on y axis,
opens down (sheds water)
so just do the right half
half A = x y
hA = -x (x^2-6) = -x^3 + 6 x
dhA/dx = 0 at max = -3 x^2 + 6
x^2 = 2
x = sqrt 2
include left half
length along x axis = 2 sqrt 2
if x = sqrt 2
y = 6 - x^2 = 6-2 = 4
thank you, somebody wad telling me you had to find the derivative first but i kept getting wrong answers from it
To find the dimensions of the rectangle with the greatest possible area, we need to maximize the area of the rectangle. The area of a rectangle is calculated by multiplying its length and width.
Let's denote the length of the rectangle as L and the width of the rectangle as W. Since the base of the rectangle is on the x-axis, the length L will be equal to the x-coordinate of the upper corners of the rectangle.
The upper corners of the rectangle lie on the parabola y = 6 - x^2. So, the y-coordinate of the upper corners will be equal to the height H of the rectangle.
To find the dimensions of the rectangle with the greatest possible area, we need to find the maximum value of the area function A(L, W). The area function can be expressed as A = L * W.
Since L = x and W = H, we need to express the area function in terms of x and H.
From the given equation of the parabola, we have y = 6 - x^2. Let's solve this equation for x.
y = 6 - x^2
x^2 = 6 - y
x = √(6 - y)
Now, we can substitute x and H in terms of y into the area function A.
A = L * W = x * H = (√(6 - y)) * H
To maximize the area, we need to find the maximum value of A. To do that, we differentiate A with respect to y and equate it to zero.
dA/dy = (√(6 - y)) * dH/dy + H * d(√(6 - y))/dy
Setting dA/dy = 0 and simplifying the equation:
0 = (√(6 - y)) * dH/dy + H * (-1/2(6 - y)^(-1/2) * (-1))
√(6 - y) * dH/dy = 1/2H * √(6 - y)
Now, we can cancel √(6 - y) from both sides:
dH/dy = 1/2H
Now, we integrate both sides with respect to y:
∫ dH/H = ∫ (1/2)dy
ln(H) = (1/2)y + C
Simplifying, we get:
H = e^((1/2)y) * C
To find the value of C, we substitute the coordinates of the upper corners of the rectangle into the equation of the parabola:
y = 6 - x^2
y = 6 - (√(6 - y))^2
y = 6 - (6 - y)
2y = 6
y = 3
Now, we can substitute y = 3 into the equation H = e^((1/2)y) * C:
H = e^((1/2)3) * C
H = e^(3/2) * C
Since H is the height of the rectangle, it cannot be negative. So, we can ignore the negative value of e^(3/2).
Now, we have the width W and the height H of the rectangle:
W = H = e^(3/2) * C
To find the value of C, we can use the fact that the rectangle is inscribed with its base on the x-axis. So, the width of the rectangle is twice the x-coordinate of the upper corners.
W = 2x
Substituting W = 2x and H = e^(3/2) * C, we get:
2x = e^(3/2) * C
Solving for C, we find:
C = (2x) / e^(3/2)
Now, substituting the value of C back into the equation for H, we get:
H = e^(3/2) * [(2x) / e^(3/2)]
H = 2x
Therefore, the dimensions of the rectangle with the greatest possible area are:
Length (L) = x
Width (W) = 2x
Height (H) = 2x
Note that the actual value of x needs to be calculated by finding the maximum of the area function A = L * W. This requires further analysis or numerical methods.