Question

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 1

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

$x=y+15$ ----> equation A

The volume of the box is equal to

$V=LWH$

we have

$V=1,350\ in^3$

$L=(x-3-3)=(x-6)\ in\\W=(y-3-3)=(y-6)\ in$

$H=3\ in$

substitute in the formula of volume

$1,350=(x-6)(y-6)(3)$ ----> equation B

substitute equation A in equation B

$1,350=(y+15-6)(y-6)(3)$

solve for y

$450=(y+9)(y-6)$

$y^2-6y+9y-54=450\\y^2+3y-504=0$

Solve the quadratic equation by graphing

using a graphing tool

The solution is y=21

Find the value of x

$x=21+15=36$

therefore

The original dimensions of the piece of metal are

Length: 36 inches

Width: 21 inches