<p>A rectangular container with two square sides and an open top is to have a volume of V cubic units. Find the dimensions of the container with minimum surface area.</p>
<p>Square side: ?</p>
<p>Third side: ?</p>
<p>Let square side = b and third side = h. We want to minimize S = b^2+4bh, so substitute h = V/b^2 to obtain S = b^2+4V/b, so just one-variable (V is constant) optimization with b>0. Set S’ = 0 = 2b - 4V/b^2 = 2b^3 - 4V so b = (2V)^(1/3) and here S’’ = 2 + 8V/b^3 > 0 so indeed we have a minimum. Then the third side is h = V/b^2, etc.</p>
<p>thats exactly what i was doing.</p>
<p>so i got (2V)^1/3 for the square sides</p>
<p>and I got V/(2V)^2/3 for the third side.</p>
<p>and it is still wrong!</p>
<p>???</p>
<p>In that case the third side might be referring to the base… then V = x^2*y and S = 2x^2 + 3xy = 2x^2 + 3V/x so min at S’ = 0 = 4x - 3V/x^2 = 4x^3 - 3V = 0 or x = (3V/4)^(1/3). S’’ > 0 here as well.</p>
<p>The box has an open top. So the original method is right; it’s just that the 4bh term should be 3bh.</p>
<p>Also, in the original method it should be 2b^2 as oppose to b^2 because you are dealing with a rectangle that has two square sides.</p>