Question

A piece of cardboard measures 10 ft by 10 ft. Four equal squares of size x are removed from the corners. After removing the squares, the piece of cardboard is then folded into an open top box. What value of x would maximize its volume?

Answer 1

Answer:

The value of x would be $\frac{5}{3}$

Step-by-step explanation:

Given,

The dimension of the cardboard = 10 ft by 10 ft,

∵ After removing four equal squares of size x ( in ft ) from the corners,

The dimension of the resultant box would be,

Length = ( 10 - 2x ) ft,

Width = ( 10 - 2x ) ft,

Height = x ft,

The volume of box,

$V=(10-2x)\times (10 - 2x)\times x=x(10-2x)^2 = x(100 - 40x + 4x^2)=100x - 40x^2 + 4x^3$

Differentiating with respect to x,

$V'=100 - 80x + 12x^2$

Again differentiating with respect to x,

$V''=-80 + 24x$

For maxima or minima,

$V'=0$

$\implies 100 - 80x + 12x^2 = 0$

$\implies 3x^2 - 20x + 25=0$

By quadratic formula,

$x=\frac{20\pm \sqrt{20^2-4\times 3\times 25}}{6}$

$x=\frac{20\pm \sqrt{400 - 300}}{6}$

$x=\frac{20\pm \sqrt{100}}{6}$

$x=\frac{20\pm 10}{6}$

$\implies x = \frac{10}{6}=\frac{5}{3}\text{ or } x = 5$

For x = 5/3, V'' = negative,

While for x = 5, V'' = Positive,

Hence, the value of x would be 5/3 ft for maximising the volume.