Question 1182974
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Let x be the inside length in meters of a side of the square base; let h be the height/depth of the tank.<br>
Then the volume of the tank is<br>
{{{x^2h=4000}}} [1]<br>
Since the thickness of the concrete is 50cm = 0.5m, the square base has dimensions (x+1)(x+1)(0.5).<br>
The four sides of the tank can be viewed as four congruent rectangular solids each with dimensions (x+0.5)(h)(0.5).<br>
So the total volume of the tank is<br>
{{{V=0.5(x+1)^2+4(x+0.5)(h)(0.5)}}} [2]<br>
Solve [1] for h in terms of x and substitute in the volume formula to get the volume in terms of the single variable x.<br>
{{{h = 4000/x^2}}}<br>
{{{V=0.5(x+1)^2+4(x+0.5)(4000/x^2)(0.5)}}}
{{{V=0.5(x+1)^2+(2x+1)(4000/x^2)}}}
{{{V=0.5(x+1)^2+8000/x+4000/x^2}}}<br>
Differentiate and set the derivative equal to zero to find the length of the side of the square base that minimizes the volume of concrete.<br>
{{{dV/dx=(x+1)-8000/x^2-8000/x^3}}}<br>
{{{x^3(x+1)-8000x-8000 = 0}}}
{{{x^3(x+1)-8000(x+1) = 0}}}
{{{(x^3-8000)(x+1) = 0}}}<br>
{{{x^3=8000}}}  or  {{{x=-1}}}<br>
Clearly the negative solution makes no sense in the problem.  So<br>
{{{x^3=8000}}}
{{{x = 20}}}<br>
ANSWER: The volume of concrete to make the tank is minimized when the side length of the square base is 20m<br>