Answered

A rectangular piece of metal is 15 in longer than it is wide. Squares with sides 3 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 1350 incubed​, what were the original dimensions of the piece of​ metal? What is the original​ width? nothing in

Answer :

calculista

Answer:

The original dimensions of the piece of metal are

Length: 36 inches

Width: 21 inches

Step-by-step explanation:

Let

x ----> the original length pf the piece of metal

y ----> the original width pf the piece of metal

we know that

[tex]x=y+15[/tex] ----> equation A

The volume of the box is equal to

[tex]V=LWH[/tex]

we have

[tex]V=1,350\ in^3[/tex]

[tex]L=(x-3-3)=(x-6)\ in\\W=(y-3-3)=(y-6)\ in[/tex]

[tex]H=3\ in[/tex]

substitute in the formula of volume

[tex]1,350=(x-6)(y-6)(3)[/tex] ----> equation B

substitute equation A in equation B

[tex]1,350=(y+15-6)(y-6)(3)[/tex]

solve for y

[tex]450=(y+9)(y-6)[/tex]

[tex]y^2-6y+9y-54=450\\y^2+3y-504=0[/tex]

Solve the quadratic equation by graphing

using a graphing tool

The solution is y=21

Find the value of x

[tex]x=21+15=36[/tex]

therefore

The original dimensions of the piece of metal are

Length: 36 inches

Width: 21 inches

Other Questions