College Algebra, Chapter 4, 4.1, Section 4.1, Problem 74

Suppose that a wire 10 cm long is cut into two pieces, one of length $x$ and the other of length $10 - x$. Each piece is bent into the shape of a square.

a.) Find a function that models the total area enclosed by the two squares.

b.) Find the value of $x$ that minimizes the total area of the two squares.

a.) If $A_1 = x^2$ be the area of the square with length $x$, then $A_2 = (10 - x)^2$ is the area of the other square. Thus, the total area is $A_T = A_1 + A_2$.


$
\begin{equation}
\begin{aligned}

A_T =& x^2 + (10 - x)^2
\\
\\
A_T =& x^2 + 100 - 20x + x^2
\\
\\
A_T =& 2x^2 - 20x + 100

\end{aligned}
\end{equation}
$


b.) The function $A_T$ is a quadratic function with $a = 2$ and $b = -20$, thus, its minimum value occurs when

$\displaystyle x = \frac{-b}{2a} = \frac{-(-20)}{2(2)} = \frac{20}{4} = 5 $ cm