Optimizing Dimensions of a Square Box: Math Lecture

Optimization. Lecture In the problem statement, we are given a square box with no lid, which we can assume has a length of l and a width of w. A square box is to be manufactured so that its surface area is 192. The question asks us to determine the dimensions that give the largest volume of the box. When we are trying to find the largest of several numbers, we are solving the problem of maximizing. Since we are trying to find the maximum, we know that we can use the derivative of our function and set it equal to 0. We can then graph it on a number line and find the maximum point. However, the problem is that we do not have an equation that will maximize both volume and surface area of our rectangular prism. So we will have to do a little bit of work to find out what the optimal dimensions of our box should be. Although we do not have an equation to maximize volume or surface area, we can maximize these attributes by maximizing the amount of space between the two objects. If I call the base side x and it is a square, then those are equal to each other. And then that I'll call h —that side I'll call h. It might not be equal to x. We don't know. But the volume, then, would be x squared times h. That's going to be my primary equation because that is the equation we want maximized. That is the equation we are going to maximize because this is what we want to maximize: volume.

Now, before you take the derivative, you have two unknown variables. In math, if you have two unknown variables, you need two equations; so we're going to consult the other thing that's been mentioned: surface area. Surface area is equal to 192; and that's because that's the base. There are four xh sides to this. To solve the equation 192 = x2 + 4xh, we have to use some kind of substitution. It is your choice whether to substitute x or h. My preference is to do the problem here. I think it'll be easier, so I can subtract x squared. It's equal to four times h. Divide by 4x and you get 48 over x minus x over 4. That's equal to h. So I'm going to replace h with that in this equation. So v is equal to x squared times what I have down for h--48 over x minus x over 4. All right. The equation v equals 48x minus x cubed over 4. We need to find the maximum. First, we will take the derivative of the product 48x minus x cubed. Then we can set this equal to 0 and solve for x. Since v prime = 48 − 3·x2/4, we get that v prime is equal to 0 when x2/4 = 0. Substituting this into the equation for v prime then yields 64 = x2, or x2 = 64. Consequently, x may be either 8 or −8 but cannot be a negative value since it is not within the domain of this function. We can solve for the length of side h. Thus, when x is 8, we have 48 over 8 minus 8 over 4 is equal to h. Therefore, 6 minus 2, which gives us a length of 4. The dimensions are therefore 8 by 8 by 4. Now let`s talk about the guidelines. If possible, make a sketch of the concept and identify what is given and what you need to find. Then write down any sort of equations and use secondary quotations to reduce the primary ones into a single variable. To find the max and min values of a first derivative, you should use calculus techniques. You will omit extraneous solutions by ensuring that your solution is specific enough to provide the details of how to solve for the max and min values. As you now know, optimization is helpful in finding maximum profit, minimum cost, and using limited resources. Businesses will use optimization models to find the maximum profit, minimum cost, and the most efficient use of scarce materials, and there are examples.