The figure at the right consists of a rectangle topped by an isosceles right triangle. The area of the figure is 200 sq. feet. The minimum perimeter of the figure is ___ft.
height of rectangle = z
width of rectangle = b = base of triangle
height of triangle = h
length of side of triangle = sqrt[h^2+(.5b)^2]
perimeter = p
= 2 z + b + 2sqrt(h^2 +.25b^2)
area = A
= z b + (b/2) h = 200
yuuck
let the width of the rectangle be 2x
let the height of the rectangle by y
then sides of triangle are 2x each and its height is
√3 x
area = (1/2((2x)(√3x) + 2xy = √3 x^2 + 2xy
200 = √3 x^2 + 2xy
u = (200 - √3x^2)/(2x)
Perimeter = P
= 6x + 2y
= 6x + 2(200 - √3x^2)/(2x)
= 6x + 200/x - √3 x
dP/dx = 6 - 200/x^2 - √3
= 0 for a min of P
200/x^2 = 6-√3
x^2 = 200/(6-√3)
x = 6.8455..
y = 5.7865
so minimum P = 6x + 2y = 52.646
62.151 units
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Let h = height of rectangle
Let b = base of rectangle
Let b = height of triangle
That makes b√2 the hypotenuse of the triangle
Therefore, P = 2h + 2b + b√2
and, A = b²√2 + hb
Rearranging the equation for A, we get h = 200/b - 0.5b
Substituting h into the equation for P, we get P = 2(200/b - 0.5b) + 2b + b√2
Simplifying, P = 400/b + b + b√2
dP/db = -400/b² + 1 + √2
Finding the zeroes, b = ±12.872
By the second derivative rule and the fact that the base cannot be negative, and, using 800/b³ as the second derivative, we find that there is a minimum when b = 12.872.
Substituting 12.872 back into the equation for P, we find that 62.151 is the minimum perimeter.
To find the minimum perimeter of the figure, we need to determine the dimensions of the rectangle and isosceles right triangle.
Let the width of the rectangle be "w" and the height of the rectangle be "h". Since the area of the figure is 200 sq. feet, we have:
Area of rectangle = w * h
Area of triangle = (1/2) * (base)^2
Since the triangle is isosceles, the base of the triangle is also "w".
We can now set up the equation to solve for the values of "w" and "h":
w * h + (1/2) * w^2 = 200
Simplifying the equation, we have:
w^2 + 2hw - 400 = 0
This is a quadratic equation, which we can solve using the quadratic formula:
w = (-2h ± √(4h^2 + 1600)) / 2
w = -h ± √(h^2 + 400)
Since we are looking for the minimum perimeter, we want the positive value of "w". Therefore, we can ignore the negative sign in front of the square root.
Now, let's consider the perimeter of the figure:
Perimeter = 2 * (w + h + w)
Perimeter = 4w + 2h
Substituting the positive value of "w" into the perimeter equation, we have:
Perimeter = 4 * (√(h^2 + 400)) + 2h
Perimeter = 4√(h^2 + 400) + 2h
To find the minimum perimeter, we need to minimize this expression.
Since the square root function is non-decreasing, the expression will be minimized when the value inside the square root, (h^2 + 400), is minimized.
To minimize (h^2 + 400), we need to set h = 0.
Plugging in h = 0 into the expression for the perimeter, we get:
Perimeter = 4√(0^2 + 400) + 2*0
Perimeter = 4√400
Perimeter = 4 * 20
Perimeter = 80 ft
Therefore, the minimum perimeter of the figure is 80 ft.