Mr.Aquino wants to paint the ceiling of a room that has length of (c^2+2cd+d^2) meters and width of (c+d) meters. If one can of paint will cover (c+d)^2 square meter, what is the minimum number of cans of paint needed? Express your answer as a polynomials.
Note that (c^2+2cd+d^2) = (c+d)^2
That means the area is (c+d)^2 * (c+d) = (c^2)^3
Now, divide the total area by the area covered by one can to get the number of cans needed.
I hope you caught my top. The total area is (c+d)^3
thanks a lot, because you help many people on how to solve that questions.
To find the minimum number of cans of paint needed, we need to determine the total area of the ceiling and then divide it by the area covered by each can of paint.
The area of the ceiling can be calculated by multiplying the length and width of the room:
Area = Length * Width
= (c^2 + 2cd + d^2) * (c + d)
= c^3 + 2c^2d + cd^2 + c^2d + 2cd^2 + d^3
= c^3 + 3c^2d + 3cd^2 + d^3.
Now, let's calculate the area covered by each can of paint:
Area covered by each can = (c + d)^2
= c^2 + 2cd + d^2.
Finally, we can find the minimum number of cans of paint needed by dividing the total area of the ceiling by the area covered by each can:
Minimum number of cans = Area / Area covered by each can
= (c^3 + 3c^2d + 3cd^2 + d^3) / (c^2 + 2cd + d^2).
Hence, the minimum number of cans of paint needed is (c^3 + 3c^2d + 3cd^2 + d^3) / (c^2 + 2cd + d^2), which is a polynomial in terms of c and d.