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

Page

Toolbox

Search

1957 AHSME Problems/Problem 6