Find the dimensions of the rectangle with area 225 square inches that has a minimum perimeter, and then find the minimum perimeter.
since 225 is 15^2 you could guess but anyway:
Lw = 225 so L = 225/w
p = 2 L + 2 w
p = 2 * 225/w + 2 w
dp/dw = 0 at min = 2 [ -225 ] / w^2 + 2
225 = w^2
15 = w
minimum perimeter is achieved when the rectangle is a square.
Well, if we want to find the dimensions of a rectangle with a minimum perimeter, we're going to have to think outside the box. Or in this case, inside the rectangle.
Let's consider the factors of 225 - 1, 3, 5, 9, 15, 25, 45, and 225. To minimize the perimeter, we want to find the pair with the smallest sum.
Upon careful consideration, we find that the pair 9 and 25 has the smallest sum (34). These two numbers represent the length and width of our rectangle.
So, the dimensions of the rectangle are 9 inches by 25 inches. (Who knew a 9 by 25 would be such a hit at parties?)
Now, to find the minimum perimeter, we just double the dimensions and add them up: (2 * 9) + (2 * 25) = 18 + 50 = 68 inches.
Therefore, the minimum perimeter is 68 inches. Now that's a rectangle that knows how to cut corners! Or rather, minimize them.
To find the dimensions of a rectangle with a minimum perimeter, we need to consider the relationship between the length and the width of the rectangle.
Let's assume that the length of the rectangle is represented by "L" and the width by "W". We know that the area of the rectangle is given by the formula: Area = Length x Width.
Given that the area is 225 square inches, we can set up the equation as:
225 = L x W
To find the dimensions that result in a minimum perimeter, we need to minimize the function representing the perimeter of the rectangle. The perimeter of a rectangle is given by the formula: Perimeter = 2L + 2W.
We can rewrite the perimeter formula as: Perimeter = 2L + 2(225/L)
To find the minimum perimeter, we need to find the value of L that minimizes the perimeter function. We can do this by taking the derivative of the perimeter function with respect to L and setting it equal to zero.
Let's differentiate the perimeter function:
d(Perimeter)/dL = 2 - (450/L^2)
Setting the derivative equal to zero:
2 - 450/L^2 = 0
Simplifying the equation:
450/L^2 = 2
Cross-multiplying:
450 = 2L^2
Dividing by 2:
225 = L^2
Taking the square root:
L = 15
Now that we have the value of L, we can substitute it back into the equation for the area to find the corresponding width:
225 = 15 x W
W = 225/15
W = 15
Therefore, the dimensions of the rectangle with an area of 225 square inches and a minimum perimeter are:
Length (L) = 15 inches
Width (W) = 15 inches
To find the minimum perimeter, we substitute the dimensions back into the perimeter formula:
Perimeter = 2L + 2W
Perimeter = 2(15) + 2(15)
Perimeter = 30 + 30
Perimeter = 60 inches
Therefore, the minimum perimeter for the rectangle is 60 inches.
the rectangle with max area per perimeter is a square
L * W = 225 ... L = 225 / W
p = 2 (L + W) = 2 [(225 / W) + W] = (450 / W) + 2 W
dp/dW = (-450 / W^2) + 2 ... the 1st derivative is zero at the minimum p
(-450 / W^2) + 2 = 0 ... (-450 / W^2) = - 2 ... 225 = W^2 ... W = 15