5(w-10)(w+15-10) = 1250
Now just solve for w. length=w+15.
Now just solve for w. length=w+15.
Let's denote the width of the cardboard as "W" (in inches).
According to the problem, the length of the cardboard is 15 inches longer than the width. Therefore, the length will be W + 15 inches.
After cutting 5 inches from each corner and folding up the remaining flaps, the dimensions of the resulting box will be as follows:
- The height of the box will be the 5-inch cut-off portion.
- The length of the box will be the remaining width of the cardboard after the cuts.
- The width of the box will be the remaining length of the cardboard after the cuts.
The volume of the box is given as 1250 cubic inches. Therefore, we have the equation:
Volume = Length × Width × Height = 1250.
Substituting the known values into the equation, we get:
(W + 15) × (W - 10) × 5 = 1250.
Simplifying further, we have:
(W + 15) × (W - 10) = 250.
Expanding the equation:
W^2 + 5W - 150 = 250.
Bringing all terms to one side:
W^2 + 5W - 150 - 250 = 0.
Simplifying further:
W^2 + 5W - 400 = 0.
Now we can solve this quadratic equation for W using factoring, completing the square, or using the quadratic formula.
Factoring, we have:
(W - 15)(W + 20) = 0.
So, either W - 15 = 0 or W + 20 = 0.
If W - 15 = 0, then W = 15 inches (width cannot be negative).
If W + 20 = 0, then W = -20 inches.
Since the width cannot be negative, we discard this solution.
Therefore, the width of the piece of cardboard is W = 15 inches.
The length of the cardboard can be found by adding 15 inches to the width:
Length = W + 15 = 15 + 15 = 30 inches.
Hence, the dimensions of the piece of cardboard are 15 inches by 30 inches.