Question 1149783
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If x is the side length of the base square and h is the height of the box, then the area of the open box is

    A(x,h) = {{{x^2}}} + {{{4xh}}}.    (1)


The volume of the box is

    V(x,h) = {{{x^2*h}}}.        (2)


We need to minimize the surface area (1) under the restriction

    {{{x^2*h}}} = 62.5  cm^3.     (3)


From (3), express  h = {{{62.5/x^2}}} and substitute it into (1). You will get then

    A(x) = {{{x^2}}} + {{{(4*62.5*x)/x^2}}} = {{{x^2}}} + {{{250/x}}}.    (4).


Thus we need to minimize function A(x) of one variable "x" expressed by (4).


For it, take the derivative


    A'(x) = 2x - {{{250/x^2}}}


and equate it to 0 (zero).  You will get


    2x - {{{250/x^2}}} = 0,

    2x^3 - 250 = 0,

    x^3 = 250/2 = 125,

    x = {{{root(3,125)}}} = 5  cm.


Then  h = {{{62.5/x^2}}} = {{{62.5/5^2}}} = 2.5 cm.


Thus the optimal dimensions are  x = 5 cm;  h = 2.5 cm.                                     <U>ANSWER</U>

The surface area of the open box is  A = {{{x^2}}} + 4*x*h = {{{5^2}}} + 4*5*2.5 = 25 + 50 = 75 cm^2.    <U>ANSWER</U>


<U>Partial CHECK</U>.  The volume = {{{x^2*h}}} = {{{5^2*2.5}}} = 62.5 cm^3.    ! Correct !
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Solved.