Question 1113459
The first (x) length makes a square that has area {{{ A[1] =  (x/4)^2 }}}
The 2nd length is 10-x, so it makes a square that has area {{{ A[2] = ((10-x)/4)^2 }}}

{{{ A[1] + A[2] =  (x/4)^2 + ((10-x)/4)^2 =  (x^2 + (100-20x+x^2))/16  = highlight( (x^2 - 10x + 50)/8 ) }}}

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I will use Calculus to find the minimum:

First we take the 1st derivative of A with respect to x (giving us the rate of change of A with respect to x):
 {{{ (dA/dx) = (2x-10)/8 }}}  —>  This is zero when x = 5cm.   So 5cm is a critical point.

The 2nd derivative of A with respect to x tells us if the function is curved up or curved down (at x=5):
 {{{ (d^2A/dx^2) = 2 }}}       so the curve is curved up (concave up) everywhere which means the critical point is at a minimum.

{{{ highlight( 5cm ) }}} minimizes the area. 
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Since this problem was multiple choice, the endpoints x=0 and x=10  need not be considered.   In solving min/max problems in general, one must consider the function value at the endpoints in addition to at all critical points found.  
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To figure out the minimum without Calculus, one way to see the minimum is to plot the graph of the function (x along the x-axis,  A(x) along the y axis).  Since this was multiple choice, you'd plot (compute A(x)) for the 4 values given, and compare their heights.  When plotting a function in general you would take use x values and plot more A(x) values.