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 09:06, 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 :
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 | |
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All AHSME Problems and Solutions |
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