A rectangular bulding must be designed in such a way that the heat loss is minim
ID: 2882512 • Letter: A
Question
A rectangular bulding must be designed in such a way that the heat loss is minimized. A different construction material is used on each side and it is known that the lateral walls lose heat at a rate of 10 units/m^2, the front and back walls at a rate of 8 units/m^2, the floor at a rate of 1 unit/m^2 and the roof at a rate of 5 units/m^2. The length and width of the bulding must be at least 30 m each, the height must be at least 4m and the volume must be exactly 4000 m^3.
1. Can another bulding be designed with less heat loss if the restrictions of wall lengths are eliminated?
Explanation / Answer
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 }.
This is the region bounded from below by the horizontal line segment from (30, 30) to ( 100/3 , 30) (let us call this line L1), from the right by the portion of the hyperbola y = 1000 x from (30, 100/3 ) to ( 100/3 , 30) (we call this curve L2) and from the left by the vertical line segment from (30, 30) to (30, 100/3 ) (denote this by L3).
2. Hx = 6y 80000x^2 , hy = 6x 64000y^2 . hx = 0 implies 6x^2 y = 80000, or, y = 80000/(6x^2 ). Substituting this in to hy = 0 gives 6x = 64000 (6x^2/80, 000)^2 ,
so x 25.54, y 20.43.
We observe that this critical point is not in D. Nest we check the boundary of D
• On L1: y = 30, h(x, 30 = 180x + 80,000/x + 6400/3 , 30 x 100/3 . Since h'(x, 30) = 180 80, 000/x2 > 0 for 30 x 100/3 ,
h(x, 30) is an increasing function with minimum h(30,30) = 10, 200 and maximum h( 100/3 , 30) 10, 533. •
On L2: y = 1000/x , h(x, 1000/x ) = 6000+64x+ 80,000/x , 30 x 100/3 .
Since h' (x, 1000/x ) = 64 80, 000/x2 < 0 for 30 x 100/3 , h(x, 1000/x) is a decreasing function with minimum h( 100/3 , 30) 10, 533 and maximum h(30, 100/3 ) 10, 587. •
On L3: x = 30, h(30, y) = 180y + 64, 000/y + 8000/3, 30 y 100/3 .
h'(30, y) = 180 64, 000/y2 > 0 for 30 y 100/3 , so h(30, y) is an increasing function of y with minimum h(30, 30) = 10, 200 and maximum h(30, 100/3 ) 10, 587
Thus the absolute minimum of 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.
the only critical point of h, which gives a local and absolute minimum is approximately h(25.54, 20.43) 9396. So a building of volume 4000 m3 with dimensions x 25.54 m, y = 20.43 m, z 4000/((25.54)(20.43)) 7.67 m has the least amount of heat loss.