Maximizing the Volume of a Box

If x was really small like 11000 of an inch you would only be folding the edges of the box up 11000 of an inch. What is the maximum volume of such a box.

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Maximizing the Volume of a Box A box with an open top is to be constructed from a square piece of cardboard 8 in.

. Here the optimal dimensions of the box are and if we put two such boxes. So if we cut X inches from our rectangle. So x y 2 since x and y are non-negative.

Starting with a rectangular sheet of metal we find the dimensions of the squares cut from each corner to give a box with maximum volume. 3x 2 12 0. Write the resulting function below.

So the volume as a function of x is given by V x x 25 - 2 x 20 - 2 x. We get an equation which can be set equal to zero. By the way solutions to this sort of problem are typically nice in some sense.

A square will be cut out from each corner of the cardboard and the sides will be turned up to form the box. Maximize the Volume of a Box. Start with a spreadsheet formula to calculate the volume of a box from a variation of length times width times depth.

We will let the variable x equal the length and width of the small square cut from each corner. If x 0 or y 0 then V x y 0 clearly the minimum. Worked solution to this question on differentiation - maximum volume of a box.

Embed the calculation as a row in a spreadsheet in which the calculation is repeated through many rows to reveal how the volume of the box varies with x the length of the square cutout noted in the above summary. Flight of a Rocket. Maximize h w d subject to 2 h w h d A w a l l w d A f l r α h w β γ d w δ.

After cutting out the squares from the corners the width of the open-top box will be 5 2 x 5-2x 5 2 x and the length will be 7 2 x 7-2x 7 2 x. If you could describe the volume of the box in terms of only one variable you could then graph the volume using Graphmatica. The length and width of the bottom of the box are both smaller than the cardboard because of the cut out corners.

Let the length and width be denoted by LW The Volume V xL-2xW-2 x For convenience let 2x u rightarrow 2V u u-Lu-W Differentiate wrt. A box with an open top is to be constructed from a square piece of cardboard 3 ft wide by cutting out a square from each of the four corners and bending up the sides_ Find the largest volume that such a box can have a Draw several diagrams to illustrate the situation some short boxes with large bases and some tall boxes with small bases_ Find. For example if the base of the box is 25 sq ft and the side of the box ortogonal to it is 4 feet long then the box volume is 25 x 4 100 cu ft.

Suppose then we want to know when the volume will be 400 cubic inches. Alternatively the area of one side of a box might be given and the height relative to that side. We must multiply the width by the length by the height.

What is the maximum volume of such a box. In this example we maximize the shape of a box with height h width w and depth w with limits on the wall area 2 h w h d and the floor area w d subject to bounds on the aspect ratios h w and w d. Find the dimensions that yield the maximum volume.

Round your answer to two decimal places in3. Wide by cutting out a square from each of the four corners and bending up the sides. Maximizing the Volume of a Box A box with an open top is to be constructed from a square piece of cardboard 18 in.

A Show that the volume V cm 3 of the brick is given by. The graph of this function is shown in the upper right corner. Embed the calculation as a row in a spreadsheet in which the calculation is repeated through many rows to reveal how the volume of the box varies with x the length of the square cutout noted in the above summary.

V 200x - 4x 3 3. Because the length and width equal 30 2 h a height of 5 inches gives a length and width of 30 2 5 or 20 inches. Start with a spreadsheet formula to calculate the volume of a box from a variation of length times width times depth.

Check that it matches the. Round your answer to two decimal places 14. The optimization problem is.

You can now replace y in your equation for the volume of the box by the formula you got in the previous question. V 15-2X 25-2XX. Figure 4 shows a solid brick in the shape of a cuboid measuring 2x cm by x cm by y cm.

Wide by cutting out a square from each of the four corners and bending up the sides. U and simplify 3u2-2uLWLW0 Solve the quadratic equation u2x frac LWmp sqrtL2W2-LW2 Now insert the given values L8 W5 2x 103 x 15. In that case just multiply the area by the height.

Using Calculus to maximize this volume we find that the maximum volume is obtained when x equals 303. So youd get a very wide shallow box. The process of finding the conditions for a function that represents a certain amount of required effort or which represents a certain needed benefit that gives minimum and maximum values is called optimization.

Exploring Optimization Using Calculus. Consider the following problem. Now that we have assigned a value for the sides we can write the equation of the volume of the box.

Youve got your answer. The total surface area of the brick is 600 cm 2. A height of 5 inches produces the box with maximum volume 2000 cubic inches.

Thus the dimensions of the desired box are 5 inches by 20 inches by 20 inches. We set V400 and solve for X. In order to maximize the volume we must first assign a value to the sides of the squares we will be cuting out of the cardboard.

Maximizing the Volume of a Box An open box with a square base is to be made from a square piece of cardboard that measures 12 mathrmcm on each side. First we should find a formula for the volume of our box. Use your equation setting length plus girth equal to 120 to solve for y in terms of x.

3 rows A standard problem in a first-semester calculus course is to maximize the volume of a box made. Plugging these into the function V x y we get V max 4. Given that x can vary.

The area of the bottom would be very close to 600 but then we multiply that by the height of 11000 to get a volume of 06 cubic inches. We will call that value x for now. Were being asked to maximize the volume of a box so well use the formula for the volume of a box and substitute in the length width and height of the open-top box.

From our mathematical knowledge we know that the volume of a box is lengthwidthheight.

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