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a rectangular piece of metal is 5 in longer than its wide. Squares with sides 1 in long or cut from the four corners and the flaps are folded up word to form an open box. if the volume is 546 in^3 ​

Sagot :

semsee45

Answer:

Original piece of metal:  23 in × 28 in

Dimensions of the box: 21 in × 26 in × 1 in

Step-by-step explanation:

Given dimensions of a rectangular piece of metal:

  • Width = x in
  • Length = (x + 5) in

If squares with sides 1 in long are cut from the four corners, and the flaps are folded upwards to form an open box, 2 inches should be subtracted from the width and the length of the piece of metal.  Therefore, the dimensions of the box are:

  • Width = (x - 2) in
  • Length = (x + 3) in
  • Height = 1 in

To find an expression for the volume of the box, multiply the width by the length by the height:

[tex]\begin{aligned}\implies \sf Volume &=\sf width \times length \times height\\&= (x - 2) \times (x + 3) \times 1\\& = (x-2)(x+3)\\&=x(x+3)-2(x+3)\\&=x^2+3x-2x-6\\&=x^2+x-6\end{aligned}[/tex]

If the volume is 546 in³ then:

[tex]\begin{aligned}\sf Volume&=546\\\implies x^2+x-6&=546\\x^2+x-6-546&=546-546\\x^2+x-552&=0\\x^2+24x-23x-552&=0\\x(x+24)-23(x+24)&=0\\(x-23)(x+24)&=0\\\implies x&=23,-24\end{aligned}[/tex]

As length is positive, x = 23 only.

To determine the original dimensions of the piece of metal, substitute the found value of x into the expressions for width and length.  Therefore, the original dimensions of the piece of metal are:

  • Width = 23 in
  • Length = 23 + 5 = 28 in

To find the dimensions of the box, substitute the found value of x into the expressions for width and length.   Therefore, the dimensions of the box are:

  • Width = 23 - 2 = 21 in
  • Length = 23 + 3 = 26 in
  • Height = 1 in
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