Solved on Mar 05, 2024

A rectangular metal piece is 25 in longer than its width. Cutting 5 in squares from the corners and folding the flaps up forms a $1120 \mathrm{in}^{3}$ open box. Find the original width and length.

STEP 1

Assumptions
1. The original piece of metal is a rectangle.
2. The length of the piece of metal is 25 inches longer than its width.
3. Squares with sides of 5 inches are cut from each of the four corners.
4. The flaps formed by cutting the squares are folded up to create an open box.
5. The volume of the resulting box is $1120 \mathrm{in}^{3}$ .
6. We are looking for the original dimensions of the piece of metal.

STEP 2

Let's denote the original width of the piece of metal as $w$ inches. According to the problem, the length is 25 inches longer than the width, so the length can be expressed as $w + 25$ inches.

STEP 3

After cutting out the squares from each corner and folding the flaps, the new width of the box will be $w - 2 \times 5$ inches, since we remove 5 inches from each side of the width.

STEP 4

Similarly, the new length of the box will be $(w + 25) - 2 \times 5$ inches, since we remove 5 inches from each side of the length.

STEP 5

The height of the box will be equal to the side of the squares that were cut out, which is 5 inches.

STEP 6

The volume of a box is calculated by multiplying its length, width, and height. Therefore, the volume of the box can be expressed as:
$Volume = (New\, width) \times (New\, length) \times (Height)$

STEP 7

Substitute the expressions for the new width, new length, and height into the volume formula:
$Volume = (w - 10) \times (w + 15) \times 5$

STEP 8

Since we know the volume of the box is $1120 \mathrm{in}^{3}$ , we can set up the equation:
$1120 = (w - 10) \times (w + 15) \times 5$

STEP 9

To simplify the equation, divide both sides by 5:
$224 = (w - 10) \times (w + 15)$

STEP 10

Expand the right side of the equation by multiplying the factors:
$224 = w^2 + 15w - 10w - 150$

STEP 11

Combine like terms:
$224 = w^2 + 5w - 150$

STEP 12

Move all terms to one side to set the quadratic equation to zero:
$w^2 + 5w - 150 - 224 = 0$

STEP 13

Simplify the equation:
$w^2 + 5w - 374 = 0$

STEP 14

Now, we need to solve the quadratic equation for $w$ . This can be done by factoring, completing the square, or using the quadratic formula. In this case, we will use the quadratic formula:
$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Where $a = 1$ , $b = 5$ , and $c = -374$ .

STEP 15

Calculate the discriminant ( $b^2 - 4ac$ ):
$\Delta = 5^2 - 4 \times 1 \times (-374)$

STEP 16

Compute the discriminant:
$\Delta = 25 + 1496$

STEP 17

Simplify the discriminant:
$\Delta = 1521$

STEP 18

Since the discriminant is a perfect square, we can find exact solutions for $w$ . Substitute $a$ , $b$ , and $\Delta$ into the quadratic formula:
$w = \frac{-5 \pm \sqrt{1521}}{2 \times 1}$

STEP 19

Calculate the square root of the discriminant:
$\sqrt{1521} = 39$

STEP 20

Substitute the square root back into the quadratic formula:
$w = \frac{-5 \pm 39}{2}$

STEP 21

Find the two possible solutions for $w$ :
$w_1 = \frac{-5 + 39}{2} = \frac{34}{2} = 17$
$w_2 = \frac{-5 - 39}{2} = \frac{-44}{2} = -22$

STEP 22

Since a width cannot be negative, we discard $w_2 = -22$ and accept $w_1 = 17$ inches as the original width of the piece of metal.

STEP 23

Now that we have the original width, we can find the original length by adding 25 inches to the width:
$Original\, length = w + 25$

STEP 24

Substitute the value of $w$ into the equation for the original length:
$Original\, length = 17 + 25$

STEP 25

Calculate the original length:
$Original\, length = 42$
The original width of the piece of metal is 17 inches, and the original length is 42 inches.

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A rectangular metal piece is 25 in longer than its width. Cutting 5 in squares from the corners and folding the flaps up forms a 1120in31120 \mathrm{in}^{3}1120in3 open box. Find the original width and length.

STEP 1

STEP 2

Let's denote the original width of the piece of metal as www inches. According to the problem, the length is 25 inches longer than the width, so the length can be expressed as w+25w + 25w+25 inches.

STEP 3

After cutting out the squares from each corner and folding the flaps, the new width of the box will be w−2×5w - 2 \times 5w−2×5 inches, since we remove 5 inches from each side of the width.

STEP 4

Similarly, the new length of the box will be (w+25)−2×5(w + 25) - 2 \times 5(w+25)−2×5 inches, since we remove 5 inches from each side of the length.

STEP 5

The height of the box will be equal to the side of the squares that were cut out, which is 5 inches.

STEP 6

The volume of a box is calculated by multiplying its length, width, and height. Therefore, the volume of the box can be expressed as:Volume=(New width)×(New length)×(Height)Volume = (New\, width) \times (New\, length) \times (Height)Volume=(Newwidth)×(Newlength)×(Height)

STEP 7

Substitute the expressions for the new width, new length, and height into the volume formula:Volume=(w−10)×(w+15)×5Volume = (w - 10) \times (w + 15) \times 5Volume=(w−10)×(w+15)×5

STEP 8

Since we know the volume of the box is 1120in31120 \mathrm{in}^{3}1120in3, we can set up the equation:1120=(w−10)×(w+15)×51120 = (w - 10) \times (w + 15) \times 51120=(w−10)×(w+15)×5

STEP 9

To simplify the equation, divide both sides by 5:224=(w−10)×(w+15)224 = (w - 10) \times (w + 15)224=(w−10)×(w+15)

STEP 10

Expand the right side of the equation by multiplying the factors:224=w2+15w−10w−150224 = w^2 + 15w - 10w - 150224=w2+15w−10w−150

STEP 11

Combine like terms:224=w2+5w−150224 = w^2 + 5w - 150224=w2+5w−150

STEP 12

Move all terms to one side to set the quadratic equation to zero:w2+5w−150−224=0w^2 + 5w - 150 - 224 = 0w2+5w−150−224=0

STEP 13

Simplify the equation:w2+5w−374=0w^2 + 5w - 374 = 0w2+5w−374=0

STEP 14

STEP 15

Calculate the discriminant (b2−4acb^2 - 4acb2−4ac):Δ=52−4×1×(−374)\Delta = 5^2 - 4 \times 1 \times (-374)Δ=52−4×1×(−374)

STEP 16

Compute the discriminant:Δ=25+1496\Delta = 25 + 1496Δ=25+1496

STEP 17

Simplify the discriminant:Δ=1521\Delta = 1521Δ=1521

STEP 18

Since the discriminant is a perfect square, we can find exact solutions for www. Substitute aaa, bbb, and Δ\DeltaΔ into the quadratic formula:w=−5±15212×1w = \frac{-5 \pm \sqrt{1521}}{2 \times 1}w=2×1−5±1521​​

STEP 19

Calculate the square root of the discriminant:1521=39\sqrt{1521} = 391521​=39

STEP 20

Substitute the square root back into the quadratic formula:w=−5±392w = \frac{-5 \pm 39}{2}w=2−5±39​

STEP 21

Find the two possible solutions for www:w1=−5+392=342=17w_1 = \frac{-5 + 39}{2} = \frac{34}{2} = 17w1​=2−5+39​=234​=17w2=−5−392=−442=−22w_2 = \frac{-5 - 39}{2} = \frac{-44}{2} = -22w2​=2−5−39​=2−44​=−22

STEP 22

Since a width cannot be negative, we discard w2=−22w_2 = -22w2​=−22 and accept w1=17w_1 = 17w1​=17 inches as the original width of the piece of metal.

STEP 23

Now that we have the original width, we can find the original length by adding 25 inches to the width:Original length=w+25Original\, length = w + 25Originallength=w+25

STEP 24

Substitute the value of www into the equation for the original length:Original length=17+25Original\, length = 17 + 25Originallength=17+25

STEP 25

Calculate the original length:Original length=42Original\, length = 42Originallength=42The original width of the piece of metal is 17 inches, and the original length is 42 inches.

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A rectangular metal piece is 25 in longer than its width. Cutting 5 in squares from the corners and folding the flaps up forms a $1120 \mathrm{in}^{3}$ open box. Find the original width and length.

Let's denote the original width of the piece of metal as $w$ inches. According to the problem, the length is 25 inches longer than the width, so the length can be expressed as $w + 25$ inches.

After cutting out the squares from each corner and folding the flaps, the new width of the box will be $w - 2 \times 5$ inches, since we remove 5 inches from each side of the width.

Similarly, the new length of the box will be $(w + 25) - 2 \times 5$ inches, since we remove 5 inches from each side of the length.

The volume of a box is calculated by multiplying its length, width, and height. Therefore, the volume of the box can be expressed as:
$Volume = (New\, width) \times (New\, length) \times (Height)$

Substitute the expressions for the new width, new length, and height into the volume formula:
$Volume = (w - 10) \times (w + 15) \times 5$

Since we know the volume of the box is $1120 \mathrm{in}^{3}$ , we can set up the equation:
$1120 = (w - 10) \times (w + 15) \times 5$

To simplify the equation, divide both sides by 5:
$224 = (w - 10) \times (w + 15)$

Expand the right side of the equation by multiplying the factors:
$224 = w^2 + 15w - 10w - 150$

Combine like terms:
$224 = w^2 + 5w - 150$

Move all terms to one side to set the quadratic equation to zero:
$w^2 + 5w - 150 - 224 = 0$

Simplify the equation:
$w^2 + 5w - 374 = 0$

Calculate the discriminant ( $b^2 - 4ac$ ):
$\Delta = 5^2 - 4 \times 1 \times (-374)$

Compute the discriminant:
$\Delta = 25 + 1496$

Simplify the discriminant:
$\Delta = 1521$

Since the discriminant is a perfect square, we can find exact solutions for $w$ . Substitute $a$ , $b$ , and $\Delta$ into the quadratic formula:
$w = \frac{-5 \pm \sqrt{1521}}{2 \times 1}$

Calculate the square root of the discriminant:
$\sqrt{1521} = 39$

Substitute the square root back into the quadratic formula:
$w = \frac{-5 \pm 39}{2}$

Find the two possible solutions for $w$ :
$w_1 = \frac{-5 + 39}{2} = \frac{34}{2} = 17$
$w_2 = \frac{-5 - 39}{2} = \frac{-44}{2} = -22$

Since a width cannot be negative, we discard $w_2 = -22$ and accept $w_1 = 17$ inches as the original width of the piece of metal.

Now that we have the original width, we can find the original length by adding 25 inches to the width:
$Original\, length = w + 25$

Substitute the value of $w$ into the equation for the original length:
$Original\, length = 17 + 25$

Calculate the original length:
$Original\, length = 42$
The original width of the piece of metal is 17 inches, and the original length is 42 inches.