Optimazation math homework

[jenjen07]

I'm lost with this problem. If anyone can help at all I'd really appreciate it.

What are the dimensions of the base of the rectangular box of greatest volume that can be constucted from 100 suqare inches of cardboard if the base is to be twice as long as it is wide? Assume that the box has a top.

I tried it and I got my voume to equal 100x^2-4x^4 divided by 3x. Maybe I have the equation wrong. Thanks for your help.

Jen

 

[StatusX]

the sides of the bottom of the box are x and 2x, and the height is y, so you want to maximize the volume:

V= 2x^2 y

with the constraint that the surface area is 100:

4x^2 + 2x y + 4 x y = 100

You could solve for y in the constraint equation, plug that in for the volume and maximize, or you could use a lagrange multiplier.

 

[M98Ranger]

Hi Jen,

I can definitely help you with this problem. First, let's start by breaking down the information given in the problem. We know that we have 100 square inches of cardboard to work with and that the base of the box is twice as long as it is wide. So, we can represent the base as 2x by x (length by width). We also know that the box has a top, which means we need to account for the height of the box as well. Let's represent the height as h.

To find the volume of the box, we use the formula V = lwh. In this case, our length is 2x, our width is x, and our height is h. So, our equation becomes V = 2x * x * h, which simplifies to V = 2x^2 * h.

Now, we have to take into account the constraint that we have 100 square inches of cardboard to work with. The surface area of a rectangular box is given by SA = 2lw + 2lh + 2wh. Since we have a top, we can ignore the top and only focus on the sides. So, our equation becomes 100 = 2(2x)(x) + 2(2x)(h) + 2(x)(h). Simplifying this, we get 100 = 4x^2 + 4xh + 2xh.

Now, we can use this equation to solve for h in terms of x. First, let's factor out an h from the right side of the equation. This gives us 100 = 4x^2 + (4x + 2)h. Then, divide both sides by (4x + 2) to isolate h. This gives us h = (100 - 4x^2)/(4x + 2).

To find the maximum volume, we need to find the value of x that will give us the largest possible value for V. To do this, we can use the formula for finding the maximum value of a quadratic function, which is -b/2a. In our case, a = 4 and b = 4x + 2. So, the value of x that will give us the maximum volume is -b/2a = -(4x + 2)/(2*4) = -(4x + 2