Question

a metal warehouse, whose dimensions are shown below, needs paint. the front and back of the warehouse each have 2 rollup doors measuring 22 ft by 28 ft each. the side of the warehouse facing the parking lot has an entry door measuring 63 in by 80 in. the other side of the warehouse has no window or door. use the given information to answer the questions. each tab shows a different view of the warehouse. front - right view back - left view conversion facts for length 1 foot (ft) = 12 inches (in) 1 yard (yd) = 3 feet (ft) 1 yard (yd) = 36 inches (in) (a) assuming the roof and doors require no paint, what is the area in square feet that needs paint? (do not round any intermediate computations and give your answer as a whole number.) 2150ft² (b) the paint to be used is sold in cans. each can contains enough paint to cover 580ft². assume there is no paint yet and partial cans cannot be bought. how many cans will need to be bought in order to paint the warehouse?

Explanation:

Step1: Calculate the total area of the walls

The warehouse has two pairs of walls.

• Front - back walls:

The area of a rectangle is \(A = l\times h\). For the front and back walls (excluding doors), the height \(h = 35\) ft. The length of the front - back side is \(l = 50\) ft.
The area of the two front - back walls (without considering doors yet) is \(2\times(50\times35)=3500\) \(ft^{2}\).

• Side walls:

The length of the side walls is \(l = 43\) ft. The area of the two side walls is \(2\times(43\times35)=3010\) \(ft^{2}\).
So the total area of the four walls (without subtracting doors and door) is \(3500 + 3010=6510\) \(ft^{2}\).

Step2: Calculate the area of the doors

• Roll - up doors:

Each roll - up door has an area \(A_{door1}=22\times28 = 616\) \(ft^{2}\). There are \(2\times2 = 4\) roll - up doors (front and back). So the total area of roll - up doors is \(4\times616 = 2464\) \(ft^{2}\).

• Entry door:

The entry door has dimensions \(63\) in by \(80\) in. First, convert inches to feet. Since \(1\) ft \(= 12\) in, the dimensions are \(\frac{63}{12}=5.25\) ft and \(\frac{80}{12}=\frac{20}{3}\approx6.67\) ft. The area of the entry door \(A_{door2}=5.25\times\frac{20}{3}=35\) \(ft^{2}\).

Step3: Calculate the area to be painted

The area to be painted \(A=6510-(2464 + 35)=4011\) \(ft^{2}\).

Step4: Calculate the number of paint cans

Each can covers \(580\) \(ft^{2}\). The number of cans \(n=\frac{4011}{580}\approx6.92\). Since partial cans cannot be bought, we use the ceiling function. So \(n = 7\).

Answer:

(a) The area that needs paint is \(4011\) \(ft^{2}\).
(b) The number of cans that need to be bought is \(7\).