document.write( "Question 320895: I have a two part question.
\n" ); document.write( "A box without a top is constructed from a 4m by 8m piece of cardboard by cutting a square of side length x from each corner.\r
\n" ); document.write( "\n" ); document.write( "a. Find the volume (v) of this box as a function of x.\r
\n" ); document.write( "\n" ); document.write( "b. By sketching the graph of the function v(x), establish that v(x) attains a maximum value on the interval (0,2). \n" ); document.write( "

Algebra.Com's Answer #229817 by toidayma(44) $\"\"$ $\"About$

You can put this solution on YOUR website!
after cutting for squares of side length x each corner, we have a box with height of x and base of (4-2x)(8-2x)
\n" ); document.write( "Of course, 0 <= x <= 2.
\n" ); document.write( "a) So, the Volume of the box is V(x) = x(4-2x)(8-2x)
\n" ); document.write( "b) the point is that to find the maximum of V(x) with x belonging to (0,2).
\n" ); document.write( "V(x) = x(4-2x)(8-2x) = 4(x^3 - 4x^2 - 2x + 8)
\n" ); document.write( "V'(x) = 4(3x^2 - 8x - 2), V'(x) = 0 <-> x1 = (8+sqrt(44/3))/6 or x2 = (8-sqrt(44/3))/6.
\n" ); document.write( "I hope that you can easily get here because it is just about solving the quadratic equation.
\n" ); document.write( "OK, by sketching the graph of V(x), you will easily see that V(x): increasing from x =0 to x = x2, then decreasing from x = x2 to x = 2. Briefly, V(x) get the maximum value at x =x2. (This value of x2 also belongs to (0,2), but x1 is larger than 2.)
\n" ); document.write( "I hope that this can guide you to the final answer of this problem. It is just one popular problem :-) \n" ); document.write( "

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