Difference between revisions of "1957 AHSME Problems/Problem 6"
Line 1: | Line 1: | ||
+ | ==Problem== | ||
+ | An open box is constructed by starting with a rectangular sheet of metal <math>10</math> in. by <math>14</math> in. and cutting a square of side <math>x</math> inches from each corner. The resulting projections are folded up and the seams welded. The volume of the resulting box is: | ||
+ | |||
+ | <math>\textbf{(A)}\ 140x - 48x^2 + 4x^3 \qquad \textbf{(B)}\ 140x + 48x^2 + 4x^3\qquad \\ \textbf{(C)}\ 140x+24x^2+x^3\qquad \textbf{(D)}\ 140x-24x^2+x^3\qquad \textbf{(E)}\ \text{none of these}</math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
The resulting metal piece looks something like this where the white parts are squares of length <math>x</math>: | The resulting metal piece looks something like this where the white parts are squares of length <math>x</math>: | ||
Latest revision as of 01:55, 19 June 2019
Problem
An open box is constructed by starting with a rectangular sheet of metal in. by in. and cutting a square of side inches from each corner. The resulting projections are folded up and the seams welded. The volume of the resulting box is:
Solution
The resulting metal piece looks something like this where the white parts are squares of length :
From here, try to visualize the rectangular prism coming together and realize the height is , the length is , and the width is . Therefore, the volume is .
1957 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.