A rectangular building with a square front is to be constructed of materials that costs 11 dollars per ft2 for the flat roof, 14 dollars per ft2 for the sides and the back, and 15 dollars per ft2 for the glass front. We will ignore the bottom of the building. If the volume of the building is 5,600 ft3, what dimensions will minimize the cost of materials? (Round your answers to the nearest hundreth such that the dimensions increase from the smallest to the largest.)

y = 23.14 ft

x = 15.56 ft

so the dimensions of the building will be = 15.56 ft x 15.56 ft x 23.14 ft

Step-by-step explanation:

the area of the roof = xy

cost of the roof = 11xy

the area of the sides and back = 2xy + x²

cost of the sides and back = 14(2xy + x²) = 28xy + 14x²

area of the glass front = x²

cost of the glass front = 15x²

minimize:

total cost = 11xy + 28xy + 14x² + 15x² = 39xy + 29x²

total volume = x²y = 5600 ⇒ constraint

39y + 58x = 2λxy

39x = λx2

x²y = 5600

λx = 39 which we replace into the first equation

39y + 58x = 78y

58x = 39y

(58x)²y = 58² · 5600

now we replace 58x by 39y

(39y)²y = 58² · 5600

39² · y³ = 58² · 5600

y³ = (58² · 5600) / 39² = 12,385.53583

y = ∛12,385.53583  = 23.14 ft

x² = 5600/ 23.14 = 242.04

x = 15.56 ft

c

step-by-step explanation:

hello, sorry being so rude after all that fighting i found it kind of funny, so as me saying sorry here is an answer!

since the table showed her that   2km = 2000m.

to determine, how many km in 2000m, she would divide by 1000.

= 2000 / 1000 = 2 km.

similarly, for 8000m = 8000 / 1000 = 8 km

cheers to savings

~

your pal papaguy