A box for shopping roofing nails must have a volume of 84 cubic feet. If the box must be 3 feet wide and its height must be 3 feet less than its length, what should the dimensions of the box be?
-
Let's set up an equation for this where "x" is the length of the box. As you know, the volume of a cube is length*width*height.
Set up an equation like this:
x(x-3)(3)=84
Divide by 3
x(x-3)=28
Mulitply the monomial by the binomial
x^2-3x=28
Subtract 28 from both sides:
x^2-3x-28=0
Factor:
(x-7)(x+4)=0
x=7 or x=-4
Since a length cannot be negative, the only answer is x=7.
You can double check by plugging in "x" in the original equation: x(x-3)(3)=84
(7)(7-3)(3)=84
(7)(4)(3)=84
84=84
Length=7 feet
Width=3 feet
Height=4 feet
Set up an equation like this:
x(x-3)(3)=84
Divide by 3
x(x-3)=28
Mulitply the monomial by the binomial
x^2-3x=28
Subtract 28 from both sides:
x^2-3x-28=0
Factor:
(x-7)(x+4)=0
x=7 or x=-4
Since a length cannot be negative, the only answer is x=7.
You can double check by plugging in "x" in the original equation: x(x-3)(3)=84
(7)(7-3)(3)=84
(7)(4)(3)=84
84=84
Length=7 feet
Width=3 feet
Height=4 feet
-
1) To find the height and length of the box, divide 84 by 3 to get 28.
2) Now set up an equation. Let's say h = height and L = length. The problem tells us that the height must be 3 feet less than its length so h = L - 3
3) We know from what we found in step number 1 that the product of the height and length should be 28. So hL = 28.
4) Substitute (L - 3) for h to get (L -3)(L) = 28
5) Simplify to get L^2 - 3L = 28
6) Move everything to one side to get: L^2 - 3L - 28 = 0. This is simply a quadratic so you can factor it to be (L - 7)(L + 4) = 0
7) L is either 7 or -4. However, a box can never have a negative dimension, so L = 7.
8) Plug that into the equation we set up in step 2 to get: h = 7 - 3 = 4.
So width is 3 feet, height is 4 feet, and length is 7 feet
2) Now set up an equation. Let's say h = height and L = length. The problem tells us that the height must be 3 feet less than its length so h = L - 3
3) We know from what we found in step number 1 that the product of the height and length should be 28. So hL = 28.
4) Substitute (L - 3) for h to get (L -3)(L) = 28
5) Simplify to get L^2 - 3L = 28
6) Move everything to one side to get: L^2 - 3L - 28 = 0. This is simply a quadratic so you can factor it to be (L - 7)(L + 4) = 0
7) L is either 7 or -4. However, a box can never have a negative dimension, so L = 7.
8) Plug that into the equation we set up in step 2 to get: h = 7 - 3 = 4.
So width is 3 feet, height is 4 feet, and length is 7 feet
-
The three dimensions are length, width, and height, or l, w, and h. You are given w = 3. If the height is 3 ft less than the length, you must have h = l - 3. The volume is V = l*w*h. So you just need to solve for the only unknown, l:
V = l*w*h = l*3*(l - 3) = 3l^2 - 9l = 84
3l^2 - 9l = 84
3l^2 - 9l - 84 = 0
Now, factor the left hand side.
3(l^2 - 3l - 28) = 0
3(l - 7)(l + 4) = 0
l - 7 = 0 or l + 4 = 0
l = 7 or -4
But since the length can't be negative, it must be l = 7
l = 7
w = 3
h = l - 3 = 7 - 3 = 4
So the box is 7 ft by 3 ft by 4 ft.
V = l*w*h = l*3*(l - 3) = 3l^2 - 9l = 84
3l^2 - 9l = 84
3l^2 - 9l - 84 = 0
Now, factor the left hand side.
3(l^2 - 3l - 28) = 0
3(l - 7)(l + 4) = 0
l - 7 = 0 or l + 4 = 0
l = 7 or -4
But since the length can't be negative, it must be l = 7
l = 7
w = 3
h = l - 3 = 7 - 3 = 4
So the box is 7 ft by 3 ft by 4 ft.
-
WLH = V
V = 84
W = 3
L = L
H = L - 3
3L(L - 3) = 84
3L^2 - 9L - 84 = 0
L^2 - 3L - 28 = 0
(L + 4)(L - 7) = 0
L = {-4, 7} ignroe negative solution
W = 3 ft
L = 7 ft
H = 4 ft
V = 84
W = 3
L = L
H = L - 3
3L(L - 3) = 84
3L^2 - 9L - 84 = 0
L^2 - 3L - 28 = 0
(L + 4)(L - 7) = 0
L = {-4, 7} ignroe negative solution
W = 3 ft
L = 7 ft
H = 4 ft
-
this should be easy cuz a cube has equal sides...duh