Question

a wire 10cm 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 square. find the value of x that minimizes the total rea of the two squres

Answer 1

Answer:

The value of x that minimizes the total area of the two squares is x = 5.

Step-by-step explanation:

A wire 10 centimeters long is cut into two pieces, with one piece measuring x centimeters and the other (10 - x) centimeters.

Each piece is bend into the shape of a square, and we want to find the value of x that minimizes the total area of the two squares.

If x is bend into a square with four equivalent sides, then each side measure:

$\displaystyle A = \frac{x}{4}$

Likewise, each side of the second square will measure:

$\displaystyle A = \frac{10-x}{4}$

The area of a square is its side squared. Hence, the area of both squares will be given by:

$\displaystyle T = \left( \frac{x}{4}\right) ^2 + \left(\frac{10-x}{4}\right)^2$

Simplify:

$\displaystyle \begin{aligned} T &= \frac{x^2}{16} + \frac{x^2 - 20x + 100}{16} \\ \\ &=\frac{2x^2-20x+100}{16}\\ \\ &= \frac{1}{8}x^2-\frac{5}{4}x+\frac{25}{4} \end{aligned}$

Since this is a quadratic with a positive leading coefficient, the minimum value will occur at the vertex point.

Recall that the x-coordinate of the vertex is given by:

$\displaystyle x = -\frac{b}{2a}$

In this case, a = 1/8 and b = - 5/4. Hence:

$\displaystyle x = -\frac{\left(-\dfrac{5}{4}\right)}{2\left(\dfrac{1}{8}\right)} = 5$

In conclusion, the value of x that minimizes the total area of the two squares is x = 5.