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A rectangular piece of metal is 25 in longer than it is wide Squares with sides 5 in long are cut from the four corners and the flaps are folded upward to form an openbox. If the volume of the box is 1530 in. What were the original dimensions of the piece of metal?What is the original width? ____in

A Rectangular Piece Of Metal Is 25 In Longer Than It Is Wide Squares With Sides 5 In Long Are Cut From The Four Corners And The Flaps Are Folded Upward To Form class=

Sagot :

Given:-

A rectangular piece of metal is 25 in longer than it is wide Squares with sides 5 in long are cut from the four corners and the flaps are folded upward to form an open box. If the volume of the box is 1530 in.

To find the original width.

The formula for volume of the box is,

[tex]v=l\times b\times h[/tex]

The volume given is 1530.

Also we have,

[tex]L=W+25,H=5[/tex]

Substiutiting the values. we get,

[tex]\left(W+15\right)*W-10*5=1530[/tex]

Divide both sides by 5,

[tex]\left(W+15\right)*W-10=306[/tex]

So,

[tex]w^2+5w-150=306[/tex]

So now we get,

[tex]w^2+5w-456=0[/tex]

Now we solve the quadratic equation,

[tex]\begin{gathered} w=\frac{-5\pm\sqrt{5^2-4\cdot\:1\cdot\left(-456\right)}}{2\cdot\:1} \\ w=\frac{-5\pm\:43}{2\cdot\:1} \\ w=19,-24 \end{gathered}[/tex]

So now we skip the negative value and take the positve value 19.

So the required value is 19.

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