y= sqrt(81-x^2)
What length and width should the rectangle have so that its area is a maximum?
(smaller value)
(larger value)
I have no clue how to do this, so if someone could please help! thank you
To find the dimensions of a rectangle that maximize its area using the given equation, we can follow these steps:
Step 1: Understand the problem
We are given the equation y = sqrt(81 - x^2) and we want to find the dimensions of a rectangle that will maximize its area.
Step 2: Understand the approach
To maximize the area of a rectangle, we need to find the dimensions that will make the rectangle as large as possible. In this case, we can relate the width of the rectangle to the variable x and the length to the variable y.
Step 3: Express the area of the rectangle
The area of a rectangle is given by the formula: A = length * width. Since we have expressed the width as x and the length as y, the area can be expressed as A = xy.
Step 4: Substitute for y
Recall that y = sqrt(81 - x^2). Substituting this value into the area equation, we have A = x * sqrt(81 - x^2).
Step 5: Maximize the area
To find the maximum area, we need to find the critical points where the derivative of the area equation is zero. So, let's find the derivative and set it equal to zero:
dA/dx = sqrt(81 - x^2) - (x^2 / sqrt(81 - x^2)) = 0
Simplifying this equation, we get:
(81 - x^2) - x^2 = 0
81 - 2x^2 = 0
2x^2 = 81
x^2 = 81/2
x = sqrt(81/2)
Step 6: Calculate the dimensions
Now that we have the value of x, we can calculate the dimensions of the rectangle.
The width is x, and the length is y = sqrt(81 - x^2).
Evaluate x to find the smaller value and larger value.
Smaller value:
x = sqrt(81/2) approximately equals 6.02
y = sqrt(81 - (6.02)^2) approximately equals 7.99
Larger value:
x = -sqrt(81/2) approximately equals -6.02 (Since the square root of a positive number can be positive or negative, both values are considered.)
y = sqrt(81 - (-6.02)^2) approximately equals 7.99
Therefore,
The smaller value for the width and length of the rectangle is approximately 6.02.
The larger value for the width and length of the rectangle is approximately 7.99.
Note: It's important to keep in mind that the dimensions of the rectangle are subject to constraints. In this case, the equation y = sqrt(81 - x^2) represents the upper half of the ellipse, so the rectangle must fit within that boundary.