Answer:
$\frac{10}{3}$ in by $\frac{14}{3}$ in by $\frac{10}{3}$ in.
Explanation:
Let's denote the length of the box as $l$, the width as $w$, and the height as $h$. We are given that the original piece of cardboard has dimensions 10 in by 12 in, so we know that:
$$l + 2h = 10 \quad \text{and} \quad w + 2h = 12.$$
To find the dimensions of the box with the largest volume, we need to maximize the volume function, which is given by:
$$V(l,w,h) = lwh.$$
Using the equations above, we can express the volume in terms of just two variables, say $l$ and $h$:
$$V(l,h) = lwh = l(10-2h)h = 10h^2 - 2h^3.$$
Now we can use calculus techniques to find the maximum of this function. To do this, we need to find the critical points, which are the values of $h$ where the derivative of $V$ with respect to $h$ is zero or undefined.
Taking the derivative of $V$ with respect to $h$, we get:
$$V'(h) = 20h - 6h^2.$$
Setting this to zero to find the critical points, we get:
$$20h - 6h^2 = 0 \quad \Rightarrow \quad h(10 - 3h) = 0.$$
This equation has two solutions: $h = 0$ and $h = \frac{10}{3}$. We can check that $h=0$ corresponds to a minimum, and $h = \frac{10}{3}$ corresponds to a maximum by computing the second derivative of $V$ with respect to $h$:
$$V''(h) = 20 - 12h.$$
At $h=0$, we have $V''(0) = 20$, which is positive, so $h=0$ is a minimum. At $h=\frac{10}{3}$, we have $V''\left(\frac{10}{3}\right) = -8\frac{1}{3}$, which is negative, so $h=\frac{10}{3}$ is a maximum.
Therefore, the maximum volume is achieved when $h=\frac{10}{3}$. Using the equations we derived earlier, we can find the corresponding values of $l$ and $w$:
$$l = 10 - 2h = \frac{10}{3}, \quad w = 12 - 2h = \frac{14}{3}.$$
So the dimensions of the box with largest volume are $\frac{10}{3}$ in by $\frac{14}{3}$ in by $\frac{10}{3}$ in.