Question

A rectangular building is being designed to minimize heat loss. The east and west walls lose

Answer 1

Answer:

Check the explanation

Step-by-step explanation:

Let x be the length of the north and south walls, y the length of the east and

west walls, and z the height of the building. The heat loss is given by

h = 10(2yz) + 8(2xz) + 1(xy) + 5(xy) = 6xy + 16xz + 20yz.

The volume is 4000 m3, so xyz = 4000, and we substitute z = 4000/(xy) to obtain the

heat loss function

h(x, y) = 6xy + 80, 000/x+ 64, 000/y.

a) Since 4000/(xy)≥4, xy≤1000, i.e., y ≤1000/x. Also x ≥ 30 and y≥ 30, so the

domain of h is D= {(x, y) : x ≥30, 30≤ y≤ 1000/x}.

is the region bounded from below by the horizontal line segment from (30

,30)to ( 100/3 , 30) (let us call this line ), from the right by the portion of the hyperbola y = 1000/x from (30, 100/3 ) to ( 100/3) (we call this curve ) and from the left by the vertical line segment from (30, 30) to (30, 100/3 ) (denote this by )

b) Kindly check the solution to question B in the attached image below.

h is h(30, 30) = 10, 200 and the dimensions of the

building that minimize heat loss are walls 30 m in length and height 4000

/302 =

40

/9 = 4.44 m.

(c) From part (b), the only critical point of

h, which gives a local and absolute minimum x ≈ 25.54 m, y = 20.43 m, z ≈ 4000 /(25.54)(20.43) ≈ 7.6

I hope it helped you.

is approximately

h(25.54, 20.43) ≈ 9396. So a building of volume 4000 m3 with dimensions.