Difference between revisions of "1957 AHSME Problems/Problem 6"
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| + | ==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>: | ||
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From here, try to visualize the rectangular prism coming together and realize the height is <math>x</math>, the length is <math>14-2x</math>, and the width is <math>10-2x</math>. Therefore, the volume is <math>x(14-2x)(10-2x)=x(4x^2-48x+40)= | From here, try to visualize the rectangular prism coming together and realize the height is <math>x</math>, the length is <math>14-2x</math>, and the width is <math>10-2x</math>. Therefore, the volume is <math>x(14-2x)(10-2x)=x(4x^2-48x+40)= | ||
\boxed{\textbf{(A) } 140x - 48x^2 + 4x^3}</math>. | \boxed{\textbf{(A) } 140x - 48x^2 + 4x^3}</math>. | ||
| + | |||
| + | == See Also == | ||
| + | {{AHSME 50p box|year=1957|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 08:06, 25 July 2024
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 :
![[asy] fill((0,4)--(4,4)--(4,0)--(6,0)--(6,4)--(10,4)--(10,10)--(6,10)--(6,14)--(4,14)--(4,10)--(0,10)--cycle,grey); draw((0,0)--(14-4,0)--(10,14)--(0,14)--cycle); draw((0,0)--(0,4)--(4,4)--(4,0)--cycle); draw((10-4,0)--(10,0)--(10,4)--(6,4)--cycle); draw((0,14)--(4,14)--(4,10)--(0,10)--cycle); draw((6,14)--(6,10)--(10,10)--(10,14)--cycle); [/asy]](http://latex.artofproblemsolving.com/a/d/1/ad166c33a6db6a5e226685c3d04553075beb094c.png)
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 
.
See Also
| 1957 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
