A rectangular box(with top) has a square base. The sum of the lengths of its 12 edges is 8 feet. What?

Question

dimensions should the box have so that its surface area is as large as possible?

Answer 1

all three dimensions should be 0.66 ft. :D

Answer 2

One thing that might help you is to remember that when maximizing the area of a rectangle, the largest area will always be the product of sides with equal length. As an example, to maximize the area of a rectangle that has a perimeter of 100 feet, the 4 sides should each be 25 feet. You could make 2 sides 10 inches, and the other 2 sides 40 inches, but as you see, this will only make 400 ft^2, while sides of length 25 will give you 625 ft^2. You can see that as you get closer and closer to equal length sides, the area will get larger and larger.
So, in the case of your box, you can divide 12 into 8 and get 0.66 feet (or about 8 inches). This is a calculus problem, but easier ones like this can be solved with common sense. I would only use this to check yourself, or on a test, if showing your work is not required. But it does work, don't you see?

Answer 3

Let width= x ft Length= x+8 ft Area=84 ft^2 x(x+8)=84 x^2+8x-84=0 (x+14)(x-6)=0 x= 6, x= -14 (not possible) dimension of room= 14 ft by 6 ft

Answer 4

x = length of one edge of the base. There are 8 of these edges.

height = (8-8x)/4 = 8(1-x)/4 = 2(1-x). Height can't be negative, so 0 < x < 1

area of each base = x²
area of each side = x2(1-x) = 2x-2x²
Surface area = two bases and four sides
= 2x² + 4(2x-2x²)
= 2x² + 8x - 8x²
= -6x² + 8x

first derivative = -12x + 8
-12x + 8 = 0 ⇒ 12x = 8 ⇒ x=(2/3)ft
2(1-x) = 2/3ft
so dimensions of the box are 2/3ft by 2/3ft by 2/3ft,
or 8in by 8in by 8in

Answer 5

Make it a cubical box for max volume. Then each of the 12 edges that add up to 8 is 8/12 of a foot long.

That makes them each 2/3 foot or .66 foot in length.

Answer 6

2s^2 + 4sh = area
area equation for rectangular box with squares of sides s and height of h

8s+4h=8
the perimeter measurements
h=2 - 2s (solved perimeter eq. for h)
2s^2 + 8s - 8s^2 = area (plugged into area eq)
8s-6s^2 = area (time to take derivative)
d area/ ds = 8 - 12s (set = 0)

s=8/12 = 2/3 foot
h=2-4/3 (plugged into perimeter eq)
h = 2/3
square sides 2/3 foot each
height of box is 2/3 feet.

Answer 7

Let x be length of base edge and y=height
Total length of edges = 8x + 4y = 8
So y=2-2x

Surface area (SA) = 2x^2 + 4xy

Substitute for y

SA=2x^2 +4x(2-2x)
= 2x^2+8x-8x^2
=8x-6x^2

Maximum SA - Differentiate and set equal to 0

8-12x=0

x=2/3 = 0.67 feet

y=2-2x = 0.67 feet also

Answer 8

0.66ft

0.66ft

0.66ft