Question

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 930 incubed/ in^3, what were the original dimensions of the piece of​ metal?

Answer 1

Answer:

The original length was 41 inches and the original width was 16 inches

Step-by-step explanation:

Let

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

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

we know that

When squares with sides 5 in long are cut from the four corners and the flaps are folded upward to form an open box

The dimensions of the box are

$L=(x-10)\ in\\W=(y-10)\ in\\H=5\ in$

The volume of the box is equal to

$V=(x-10)(y-10)5$

$V=930\ in^3$

so

$930=(x-10)(y-10)5$

simplify

$186=(x-10)(y-10)$ -----> equation A

Remember that

The piece of metal is 25 in longer than it is wide

so

$x=y+25$ ----> equation B

substitute equation B in equation A

$186=(y+25-10)(y-10)$

solve for y

$186=(y+15)(y-10)\\186=y^2-10y+15y-150\\y^2+5y-336=0$

Solve the quadratic equation by graphing

using a graphing tool

The solution is y=16

see the attached figure

Find the value of x

$x=16+25=41$

therefore

The original length was 41 inches and the original width was 16 inches