Respuesta :
Answer:
Sidelengths of the box will be:
x = 4 ft, h = 2 ft.
Step-by-step explanation:
As the open rectangular box is to be made from a sheet that is 48 square feet, the surface area of this box will be at most 48 square feet.
mathematically, we can write the above information as:
[tex]\text{surface area} = 48 \text{ft}^2[/tex]
[tex]S = 48\,\text{ft}^2[/tex]
the surface area of any rectangular box can be found by adding the areas of each of its face. In our case:
- this will exclude the top face since the box is 'open'. (think of it like the lid is removed)
- the base is squared, so it side lengths are equal.
- let the lengths of the base be denoted by [tex]x[/tex]
- and height be denoted by [tex]h[/tex]
[tex]\text{surface area} = x^2 + xh + xh+xh + xh[/tex]
[tex]S = x^2 + 4xh[/tex]
now we can combine our two surface area equation
[tex]x^2 + 4xh = 48[/tex]
Now we're asked to find the dimension so that the volume is largest!
we first need to find the volume:
[tex]V = x\times x\times h[/tex]
[tex]V = x^2 h[/tex]
since to find the maximum volume we'll have to differentiate the above equation, but before that we need to have only ONE variable at each side of the equation.
We need to replace 'h'. To do that we'll use our derived equation for surface area.
[tex]x^2 + 4xh = 48[/tex]
[tex]h = \dfrac{48-x^2}{4x}[/tex]
Now we can substitute this 'h' in 'V'
[tex]V = x^2 h[/tex]
[tex]V = x^2 \left(\dfrac{48-x^2}{4x}\right)[/tex]
[tex]V = \dfrac{48x-x^3}{4}[/tex]
Now we can find the dimension at which the volume is maximum by using:[tex]\dfrac{dV}{dx}=0[/tex]
[tex]\dfrac{dV}{dx} = \dfrac{1}{4}\dfrac{d}{dx}(48x-x^3)[/tex]
[tex]\dfrac{dV}{dx} = \dfrac{1}{4}(48-3x^2)[/tex]
[tex]0 = \dfrac{1}{4}(48-3x^2)[/tex]
[tex]3x^2 = 48[/tex]
[tex]x = \sqrt{16}[/tex]
[tex]x = 4\,\text{ft}[/tex]
this the dimension of sidelength 'x' of the rectangular box such that the volume is maximum.
we can use this value of x to find the height as well.
[tex]h = \dfrac{48-(4)^2}{4(4)}[/tex]
[tex]h = 2\,\text{ft}[/tex]
at these dimensions the maximum possible volume of the box will be!
[tex]V = (4)^2 (2)[/tex]
[tex]V = 32\,\text{ft}^3[/tex]
The volume of a box is the amount of space in it.
The dimension that produces the largest possible volume is a base length of 4 feet, and a height of 2 feet.
The surface area of the box is:
[tex]\mathbf{Box = 48}[/tex]
Assume the dimension of the base is x, and the height is h.
The surface area of the box will be:
[tex]\mathbf{Box = x^2 + 4xh}[/tex]
So, we have:
[tex]\mathbf{x^2 + 4xh =48}[/tex]
Make h the subject
[tex]\mathbf{4xh =48 - x^2}[/tex]
[tex]\mathbf{h =\frac{48 - x^2}{4x}}[/tex]
The volume of the box is:
[tex]\mathbf{V = x^2h}[/tex]
So, we have:
[tex]\mathbf{V =x^2 \times \frac{48 - x^2}{4x}}[/tex]
Simplify
[tex]\mathbf{V =x \times \frac{48 - x^2}{4}}[/tex]
[tex]\mathbf{V =\frac{48x - x^3}{4}}[/tex]
Split
[tex]\mathbf{V =12x - \frac{x^3}{4}}[/tex]
Differentiate
[tex]\mathbf{V' =12 - 3 \times \frac{x^2}{4}}[/tex]
[tex]\mathbf{V' =12 - \frac{3x^2}{4}}[/tex]
Set to 0
[tex]\mathbf{12 - \frac{3x^2}{4} = 0}[/tex]
Rewrite as:
[tex]\mathbf{\frac{3x^2}{4} = 12}[/tex]
Multiply both sides by 4/3
[tex]\mathbf{x^2= 16}[/tex]
Take positive square root of both sides
[tex]\mathbf{x= 4}[/tex]
Recall that: [tex]\mathbf{h =\frac{48 - x^2}{4x}}[/tex]
So, we have:
[tex]\mathbf{h = \frac{48 - 4^2}{4 \times 4}}[/tex]
[tex]\mathbf{h = \frac{32}{16}}[/tex]
[tex]\mathbf{h = 2}[/tex]
Hence, the dimension that produces the largest possible volume is a base length of 4 feet, and a height of 2 feet.
Read more about volumes at:
https://brainly.com/question/13101997
