Answer
a) $A(x)=\frac{1}{8}x^2-\frac{5}{4}x+\frac{25}{4}$
b) $5$ cm
Work Step by Step
Formula of the area of a square: $A=x^2$, with $x$ is its side length.
a) As a cord with length $x$ is bent into a square, the total length of the cord, or the square's perimeter is still $x$. And because the sides in a square are equal in length, each side in this case is $\frac{x}{4}$ cm long. Therefore, the area of this square is $\frac{1}{16}x^2$.
Likewise, each side of the remaining square is $\frac{10-x}{4}$ cm long. That square area is:
$(\frac{10-x}{4})^2=\frac{(10-x)^2}{16}=\frac{x^2-20x+100}{16}=\frac{1}{16}x^2-\frac{5}{4}x+\frac{25}{4}$
Therefore, the function that models the total area enclosed is:
$A(x)=\frac{1}{16}x^2+\frac{1}{16}x^2-\frac{5}{4}x+\frac{25}{4}$
$A(x)=\frac{1}{8}x^2-\frac{5}{4}x+\frac{25}{4}$
(with $a=\frac{1}{8}$, $b=\frac{-5}{4}$, $c=\frac{25}{4}$)
b) We can easily apply our known formula about the minimum/maximum value in quadratic functions to answer this problem.
The minimum value occurs at:
$x=\frac{-b}{2a}=\frac{5}{4}\div(2*\frac{1}{8})=\frac{5}{4}\div\frac{1}{4}=\frac{5}{4}*4=5$
Therefore, we should cut the cord into two five-centimeter sections to minimize the area enclosed.
