Find the dimensions of a rectangle with perimeter 40 with the largest area. (hint: Find an equation for area and one for perimeter, use both to find perimeter in terms of one variable)
let the width be x
and the length be y
2x + 2y = 40 or
x+y=20 ---> y = 20-x
Area = xy = x(20-x
= 20x - x^2
Are you in Calculus?
Then the solution is very easy
If not in Calculus, complete the square of this quadratic, again a very basic problem.
Let me know what you get.
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To find the dimensions of a rectangle with the largest possible area, given its perimeter, let's start by defining the equation for the perimeter of a rectangle.
The perimeter of a rectangle is given by the formula: P = 2L + 2W, where L represents the length and W represents the width of the rectangle.
In this case, the perimeter is given as 40. So, we can write the equation as:
40 = 2L + 2W
Now, let's solve this equation for one variable in terms of the other variable.
Dividing both sides of the equation by 2, we get:
20 = L + W
Next, we can define the equation for the area of a rectangle.
The area of a rectangle is given by the formula: A = L * W.
Using the equation we derived for the perimeter (20 = L + W), we can rewrite one of the variables in terms of the other.
Substituting L = 20 - W into the equation for area, we get:
A = (20 - W) * W
Now, we have an equation for the area that depends only on the width, W.
To find the maximum area, we need to find the value of W that maximizes the equation for area (A).
To do this, we can use calculus by taking the derivative of A with respect to W and setting it equal to zero to find the critical points. We then evaluate the second derivative to confirm that the critical point is a maximum.
dA/dW = 20 - 2W
Setting dA/dW equal to zero:
20 - 2W = 0
2W = 20
W = 10
Substituting this value of W back into our equation for the perimeter:
20 = L + 10
L = 10
Therefore, the dimensions of the rectangle with the largest area and a perimeter of 40 are:
Length (L) = 10 units
Width (W) = 10 units