## How do you find the volume of a box with an open top?

The formula for volume of the box is V=l×l×h . You can determine the maximum value of this function using graphing calculator.

## How do you solve a calculus optimization problem?

To solve an optimization problem, begin by drawing a picture and introducing variables. Find an equation relating the variables. Find a function of one variable to describe the quantity that is to be minimized or maximized. Look for critical points to locate local extrema.

When is the volume of the open-top box maximized?

Because the derivative is increasing ( 1 4 > 0 14>0 1 4 > 0) to the left of the critical point, and decreasing ( − 1 3 < 0 -13<0 − 1 3 < 0) to the right of it, the function has a maximum at x = 0. 9 6 x=0.96 x = 0. 9 6, and we can say that the volume of the open-top box is maximized when x = 0. 9 6 x=0.96 x = 0. 9 6.

How can I maximize the enclosed volume of a box?

Determine the dimensions of the box that will maximize the enclosed volume. Solution We want to build a box whose base length is 6 times the base width and the box will enclose 20 in 3. The cost of the material of the sides is \$3/in 2 and the cost of the top and bottom is \$15/in 2.

### Can we use optimization in the real world?

This same optimization process can be used in the real world. When the function we start with models some real-world scenario, then finding the function’s highest and lowest values means that we’re actually finding the maximum and minimum values in that situation.

### What are the critical points of the volume function?

So the critical points of the volume function are x ≈ 3. 0 4 x\\approx3.04 x ≈ 3. 0 4 and x ≈ 0. 9 6 x\\approx0.96 x ≈ 0. 9 6. But before we start testing critical points, we should always consider which of the critical points is even plausible.