4.7 Applied optimization problems  (Page 2/8)

Step 1: Let $x$ be the side length of the square to be removed from each corner ( [link] ). Then, the remaining four flaps can be folded up to form an open-top box. Let $V$ be the volume of the resulting box.

Step 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize $V.$

Step 3: As mentioned in step $2,$ are trying to maximize the volume of a box. The volume of a box is $V=L·W·H,$ where $L,W,\text{and}H$ are the length, width, and height, respectively.

Step 4: From [link] , we see that the height of the box is $x$ inches, the length is $36-2x$ inches, and the width is $24-2x$ inches. Therefore, the volume of the box is

Step 5: To determine the domain of consideration, let’s examine [link] . Certainly, we need $x>0.$ Furthermore, the side length of the square cannot be greater than or equal to half the length of the shorter side, $24$ in.; otherwise, one of the flaps would be completely cut off. Therefore, we are trying to determine whether there is a maximum volume of the box for $x$ over the open interval $\left(0,12\right).$ Since $V$ is a continuous function over the closed interval $\left[0,12\right],$ we know $V$ will have an absolute maximum over the closed interval. Therefore, we consider $V$ over the closed interval $\left[0,12\right]$ and check whether the absolute maximum occurs at an interior point.

Step 6: Since $V\left(x\right)$ is a continuous function over the closed, bounded interval $\left[0,12\right],$ $V$ must have an absolute maximum (and an absolute minimum). Since $V\left(x\right)=0$ at the endpoints and $V\left(x\right)>0$ for $0<x<12,$ the maximum must occur at a critical point. The derivative is

To find the critical points, we need to solve the equation

Dividing both sides of this equation by $12,$ the problem simplifies to solving the equation

Using the quadratic formula, we find that the critical points are

Since $10+2\sqrt{7}$ is not in the domain of consideration, the only critical point we need to consider is $10-2\sqrt{7}.$ Therefore, the volume is maximized if we let $x=10-2\sqrt{7}\text{in}.$ The maximum volume is $V\left(10-2\sqrt{7}\right)=640+448\sqrt{7}\approx 1825\text{in}{.}^3$ as shown in the following graph.