# Difference between revisions of "1957 AHSME Problems/Problem 6"

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 $x(14-2x)(10-2x)=x(4x^2-48x+40)= \boxed{\textbf{(A) } 140x - 48x^2 + 4x^3}$.

 1957 AHSME (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 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

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