Question 1148825
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Let x be the length of the side of the base, and

let y be the height of the container.


Then the volume is  {{{x^2*y}}} = 18000 cubic inches.


The cost of the material for two bases is  {{{2*x^2}}} dollars;

the cost of the material for four lateral sides is  4*3*x*y = 12xy dollars.


The problem requires us to minimize  {{{2x^2 + 12xy}}  under the restriction   {{{x^2*y}}} = 18000.


Using the restriction formula, express  y = {{{18000/x^2}}}  and substitute it  into the total cost expression.


Then the cost takes the form  C(x) = {{{2x^2 + 12xy}}} = {{{2x^2 + (12*18000)/x}}}.


Thus the problem is just reduced to finding minimum of the function


    C(x) = {{{2x^2 + 216000/x}}}.


To find it minimum, take a derivative and equate it to zero.  It gives you the equation

    C'(x) = {{{4x - 216000/x^2}}} = 0,

which implies

    4x^3 = 216000,

     x^3 = 54000

     x = {{{root(3,54000)}}} = {{{30*root(3,2)}}}.


Then  y = {{{18000/x^2}}} = {{{18000/(900*root(3,4))}}} = {{{20/root(3,4)}}}.


<U>ANSWER</U>.  x= {{{30*root(3,2)}}} = 37.798 inches (approximately)  and  y= {{{20/root(3,4)}}} = 12.6 inches (approximately).


<U>PARTIAL CHECK</U>.  {{{x^2*y}}} = {{{37.798^2*12.6}}} = 18001 in^3.  The miserable difference is due to rounding.
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Solved.