Question

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.

Answer 1

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

Answer 2

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

Answer 3

62.151 units
------
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.