What's the value of #h# which minimize the perimeter?

Question

Consider a window the shape of which is a rectangle of height #h# surmounted by a triangle having a height #T# that is 0.9 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.

I found the value of #w# but couldn't find the value of #h#.

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Answer 1 · 2017-11-21T09:53:28

See below.

Cross sectional area

#A = h * w +1/2T*w = h*w+1/2*0.9*w^2#

Perimeter

#P = w + 2*h + 2 sqrt(T^2+(w/2)^2) = w + 2*h + 2 w sqrt(0.9^2+(1/2)^2) = 2*h + (2sqrt(0.9^2+(1/2)^2) + 1)w#

so calling #c_0 = 0.9/2, c_1 = 2, c_2 = (2sqrt(0.9^2+(1/2)^2) + 1)#

we have

find

#min P = c_1h+c_2w#

subjected to

#A = h*w+c_0 w^2#

substituting #h = (A-c_0 w^2)/w# into #P# we get

#P = c_1 w + (c_1 (A - c_0 w^2))/w#

and

#(dP)/(dw) = -2 c_0 c_1 + c_2 - (c_2 (A - c_0 w^2))/w^2 = 0#

giving

#w_0 = (sqrt[c_1A])/sqrt[c_2-c_0 c_1] = 0.962445 sqrt[A]#

and consequently

#h_0 = (A-c_0 w_0^2)/w_0 = 2sqrt(A c_1(c_2-c_0c_1)) = 0.60592 sqrt[A]#