?(H) MAA MATHEMATICAL ASSOCIATION OF AMERICA webwork / 2018-01-gedb001-54 / chap

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?(H) MAA MATHEMATICAL ASSOCIATION OF AMERICA webwork / 2018-01-gedb001-54 / chapter4-5/11 Chapter4-5: Problem 11 Previous Problem List Next 1 point) Consider a window the shape of which is a rectangle of height h surmounted by a triangle having a height T that is 1.1 times the width w of the rectangle (as shown in the figure below) If the cross-sectional area is A, determine the dimensions of the window which minimize the perimeter hi 0608w w-1 (2A12(1 46)"(1/2)+1))"(1(2) Note: You can earn partial credit on this probiem You have attempted this problem 5 times. score is 0% You have unlimilted attempts remaining

Explanation / Answer

We are given that the height of the triangular portion of the window is T and T is 1.1 times the width of the rectangular portion of the window

=> T = 1.1w   ----> (1)

The height of the rectangular portion of the window is = h

Lets first find the total area of the window

Total area , A = (Area of rectangular part) + (Area of the Triangular part)

             A = (width*height) = 1/2*(height)*(base)

             A = wh + 1/2*T*w   ----> (2)

From (1) and (2)

A = wh + 1/2*(1.1w)*w

A = wh + 0.55w^2    -----> (3)

Next we find the function for the perimeter of the window , P

=> P = (Perimeter of the rectangular part) + (Perimeter of the triangular part)

Let the slant part of the triangle be = L

=> P = (2h) + (2L)   -------> (4)

Lets use Pythagores theorem to find L

(w/2)^2 + T^2 = L^2

We know T = 1.1w

=> (w/2)^2 + (1.1w)^2 = L^2

=> w^2/4 + 1.21w^2 = L^2

=> (5.84/4)*w^2 = L^2

or L = (sqrt(5.84)/2)*w   ----> (5)

From (4) and (5)

=> P = (2h) + 2*(sqrt(5.84)/2)*w

P = 2h + sqrt(5.84)w ------> (6)

The area function given in equation (3) could be writen as:

h = (A - 0.55w^2)/w ----> (7)

From (6) and (7)

P = 2(A - 0.55w^2)/w + sqrt(5.84)w

P = (2A)/w - 1.1w + sqrt(5.84)w

or P = (2A)/w + (sqrt(5.84) - 1.1)w

Please note as the Area of the window is fixed, this implies A is constant

Lets differentiate P with respect to 'w'

=> P '(w) = -(2A)/w^2 + (sqrt(5.84) - 1.1)

next find the critical point(s) by solving p '(w) = 0

=> P '(w) = -(2A)/w^2 + (sqrt(5.84) - 1.1) = 0

w^2 = (2A)/(sqrt(5.84) - 1.1)

or w = +-sqrt[(2A)/(sqrt(5.84) - 1.1)]

we would have to neglect the negative value of w as the dimension cannot be negative

=> the critical point is :

w = sqrt[(2A)/(sqrt(5.84) - 1.1)]

Lets find the double derivative of P(w)

=> P '(w) = -(2A)/w^2 + (sqrt(5.84) - 1.1)         
   
=> p ''(w) = (4A)/w^3

As P ''(w) is positive and since we have just 1 critical point

so we have a minima at w = sqrt[(2A)/(sqrt(5.84) - 1.1)] ----> (8)

And the corresponding value of h is

From equation (7)

h = (A - 0.55w^2)/w

plug (8) in the above equation

=> h = A/w - 0.55w

or h = A/(sqrt[(2A)/(sqrt(5.84) - 1.1)]) - 0.55(sqrt[(2A)/(sqrt(5.84) - 1.1)])

Hence the dimensions that minimize the perimeter of the window are :

w = sqrt[(2A)/(sqrt(5.84) - 1.1)]

and h = A/(sqrt[(2A)/(sqrt(5.84) - 1.1)]) - 0.55(sqrt[(2A)/(sqrt(5.84) - 1.1)])

    

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