Question

1. congruent squares are cut from the corners of a 1m square piece of tin. the edges are turned up to make an open rectangular box. how large should the squares cut from the corners of the box be in order to maximize the volume of the box?

Explanation:

Step1: Define the variables

Let the side - length of the square cut from each corner be $x$ meters. The length and width of the base of the open - rectangular box will be $(1 - 2x)$ meters, and the height of the box is $x$ meters. The volume $V$ of the box is given by the function $V(x)=(1 - 2x)(1 - 2x)x=x(1 - 2x)^2$, where $0\lt x\lt0.5$.

Step2: Expand the volume function

Expand $V(x)=x(1 - 2x)^2=x(1 - 4x + 4x^{2})=x-4x^{2}+4x^{3}$.

Step3: Find the derivative

Differentiate $V(x)$ with respect to $x$. Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $V^\prime(x)=1-8x + 12x^{2}$.

Step4: Set the derivative equal to zero

Set $V^\prime(x)=0$, so $12x^{2}-8x + 1 = 0$.

Step5: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 12$, $b=-8$, $c = 1$), the quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute the values: $x=\frac{8\pm\sqrt{(-8)^{2}-4\times12\times1}}{2\times12}=\frac{8\pm\sqrt{64 - 48}}{24}=\frac{8\pm\sqrt{16}}{24}=\frac{8\pm4}{24}$.
We get two solutions: $x_1=\frac{8 + 4}{24}=\frac{12}{24}=\frac{1}{2}$ and $x_2=\frac{8 - 4}{24}=\frac{4}{24}=\frac{1}{6}$. But $x=\frac{1}{2}$ is not in the domain $0\lt x\lt0.5$.

Step6: Check the second - derivative

Differentiate $V^\prime(x)$ to get the second - derivative $V^{\prime\prime}(x)=-8 + 24x$. Substitute $x=\frac{1}{6}$ into $V^{\prime\prime}(x)$: $V^{\prime\prime}(\frac{1}{6})=-8+24\times\frac{1}{6}=-8 + 4=-4\lt0$. So the volume is maximized at $x=\frac{1}{6}$ m.

Answer:

$\frac{1}{6}$ m