Volume of a Box. A rectangular piece of metal is 10 in. longer th...

Question

Volume of a Box. A rectangular piece of metal is 10 in. longer than it is wide. Squares with sides 2 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 832 in.^3, what were the original dimensions of the piece of metal?

Accepted Answer

Hello, everyone in this video, we're going to be looking at this practice problem where we want to figure out the dimensions of a titanium bar used and were given a couple of characteristics about that titanium bar. So one of the first characteristics is that the bar has a length that is 20 cm more than be with. And when corners We're cut from this bar, they had an area of 16 cm squared. And once those squares were cut out, the titanium bar was turned into a box that had a volume of 176 cm cubed. So we want to try to find the dimensions of the Titanium bar used. So the first thing I'm going to do is draw a picture. So say this is my Titanium bar and this is the width and then I have the length. Well, we're given the characteristic that the length is 20 cm more than the width. So I'll say that the length is equal to the width plus 20. And then we're given that square corners were cut from the corner of this bar to form a box. And because they were square corners that were cut and they gave us the area, we can figure out the length here, which I'll call a that was cut. So the volume of a square is the length squared. So a times A would give us that area. So a squared is equal to 16 centimeters squared. So I take the square root, I get that A is equal to four centimeters. So that means that each of these sides is four centimeters, or each of the links of the corners is four centimeters. And then when those flops were folded upwards, the volume Was 176. Well, when we think of the volume, that's equal to the width times the length times the height. And when we fold the flaps upward, you can see that the height becomes four centimeters. But because we cut those squares out, we have to account for them in the new width and length. So you can see that the original width would have been this whole area. But the width for the volume of the box gets rid of four centimeters on the left and four centimeters on the right. So eight centimeters total. So I want to say w not the new width or the width of the box is equal to The original with -8 cm. And because Cutting the boxes also affects the length, we also have to consider that the length is also getting reduced by 8cm. So say L not the length for the box is now L minus eight centimeters. And we had already defined L as W plus 20. So I can say W plus 20 minus eight, which means that L not is equal to W positive 20-8 is positive 12. And now I can plug these into my volume formula. I have the with which is W - times the link which is W plus Times the height which is four And that is equal to 176. So in order to solve for W, I'll start by dividing this for, if I do it to one side, I have to do it to the other. So these fours cancel out and I'm left with W -8 times W plus is equal to 176 divided by four Which is 44. And then I can multiply this out. So I have W times W is W squared, W times 12 is 12, W negative eight times W is negative eight, W and the negative eight times positive is negative 96 still equal to 44. From here. You can see I can combine my middle term. So positive 12 eight is positive for W. And in order to get all my numbers on one side, I'm going to subtract by 44. So this becomes W squared plus four, W negative 96 minus 44 is negative 1 40 equal to zero. Now, you can see that this follows a quadratic equation. So I'm going to try to solve for W by factoring. So I want the terms of the first terms of my factors to multiply to get me W squared. So W times W gives me W squared. So I'll start there. Then I want the Second terms of my factors to multiply to give me -140. So when I think of factors of negative 1 40 I can do negative 14 times positive 10 or vice versa with positive 14 and negative 10 or seven and negative 20 or positive 20 and negative seven. In this case, I'm just going to try W plus 14 and W minus 10 and we can try, we can check it by doing a quick foil here where I have W times W is W squared, W times negative 10 is negative 10, W positive 14 times W is 14, W and positive 14 minus 10 times negative 10 is negative 1 40. If I combine the middle terms, I have negative 10, W plus 14 that becomes plus for W you can see that these factors give me my original equation back. So I know that they are correct, which means I can solve for W using the factors. So W plus 14 equals zero and W -10 equals zero. So I can have subtract by 14, W equals negative 14 by add 10, I get W equals positive 10. Well, in this case, it doesn't make sense for dimensions to be negative. So W equals negative 14 gets eliminated. That means that my original with was 10 centimeters. And if we look back, you can see that the original relationship was that the length equals the, with plus 20. So we plug in 10 for the, with, We get that the length is 30 cm. So this becomes my final answer for the dimensions of the original Titanium Bar. And if we look at the answer choices, You can see that answer choice. C also lists the link that's 30 cm and the width as 10cm. So hopefully, this video helped and see you next time.

My Courses

Chemistry

Biology

Math

Physics

Business

Social Sciences

Programming

Product & Marketing

Volume of a Box. A rectangular piece of metal is 10 in. longer than it is wide. Squares with sides 2 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 832 in.^3, what were the original dimensions of the piece of metal?