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Difference between revisions of "1957 AHSME Problems/Problem 6"
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==Problem==
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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:
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<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>
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==Solution==
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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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\boxed{\textbf{(A) } 140x - 48x^2 + 4x^3}</math>.
 
\boxed{\textbf{(A) } 140x - 48x^2 + 4x^3}</math>.
  
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{{AHSME box|year=1957|ab=B|num-b=5|num-a=7}}
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== See Also ==
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{{AHSME 50p box|year=1957|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{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 $10$ in. by $14$ in. and cutting a square of side $x$ inches from each corner. The resulting projections are folded up and the seams welded. The volume of the resulting box is:

$\textbf{(A)}\ 140x - 48x^2 + 4x^3 \qquad \textbf{(B)}\ 140x + 48x^2 + 4x^3\qquad \\ \textbf{(C)}\ 140x+24x^2+x^3\qquad \textbf{(D)}\ 140x-24x^2+x^3\qquad \textbf{(E)}\ \text{none of these}$

Solution

The resulting metal piece looks something like this where the white parts are squares of length $x$:

[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]

From here, try to visualize the rectangular prism coming together and realize the height is $x$, the length is $14-2x$, and the width is $10-2x$. Therefore, the volume is $x(14-2x)(10-2x)=x(4x^2-48x+40)= \boxed{\textbf{(A) } 140x - 48x^2 + 4x^3}$.

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

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