document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1192304>Question 1192304</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A cuboid has a square base of side x cm and its height is (2x + 1)cm. Find the approximate increase in its volume when x increases from 10 cm to 10.05 cm. Hence deduce an approximate value for the volume when x = 10.05 cm. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849056 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=849056\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> **1. Find the Volume Function**\r<BR>\n" );
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document.write( "* The volume (V) of the cuboid is given by:<BR>\n" );
document.write( "   V(x) = x * x * (2x + 1) <BR>\n" );
document.write( "   V(x) = 2x³ + x²\r<BR>\n" );
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document.write( "**2. Find the Initial Volume**\r<BR>\n" );
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document.write( "* When x = 10 cm:<BR>\n" );
document.write( "   V(10) = 2 * (10)³ + (10)² <BR>\n" );
document.write( "   V(10) = 2000 + 100 <BR>\n" );
document.write( "   V(10) = 2100 cm³\r<BR>\n" );
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document.write( "**3. Find the Rate of Change of Volume (dV/dx)**\r<BR>\n" );
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document.write( "* Differentiate the volume function with respect to x:<BR>\n" );
document.write( "   dV/dx = d/dx (2x³ + x²)<BR>\n" );
document.write( "   dV/dx = 6x² + 2x\r<BR>\n" );
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document.write( "**4. Approximate Increase in Volume**\r<BR>\n" );
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document.write( "* Use the differential approximation:<BR>\n" );
document.write( "   ΔV ≈ (dV/dx) * Δx <BR>\n" );
document.write( "   where ΔV is the approximate change in volume and Δx is the change in x.\r<BR>\n" );
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document.write( "* Δx = 10.05 cm - 10 cm = 0.05 cm\r<BR>\n" );
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document.write( "* ΔV ≈ (6 * (10)² + 2 * 10) * 0.05 <BR>\n" );
document.write( "   ΔV ≈ (600 + 20) * 0.05 <BR>\n" );
document.write( "   ΔV ≈ 620 * 0.05 <BR>\n" );
document.write( "   ΔV ≈ 31 cm³\r<BR>\n" );
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document.write( "**5. Approximate Volume at x = 10.05 cm**\r<BR>\n" );
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document.write( "* Approximate Volume = Initial Volume + Approximate Increase in Volume<BR>\n" );
document.write( "* Approximate Volume = 2100 cm³ + 31 cm³<BR>\n" );
document.write( "* Approximate Volume = 2131 cm³\r<BR>\n" );
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document.write( "**Therefore:**\r<BR>\n" );
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document.write( "* The approximate increase in volume when x increases from 10 cm to 10.05 cm is **31 cm³**.<BR>\n" );
document.write( "* The approximate volume of the cuboid when x = 10.05 cm is **2131 cm³**.<BR>\n" );
document.write( " </SPAN>\n" );
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