A closed box with square base is to be built. the bottom and the top of the box are to be made of a material costing $2/ft^2, and all four sides are to be made of a material costing $1/ft^2. what are the dimensions of the box of the greatest value that can be constructed for $12?
To find the dimensions of the box with the greatest value, we need to optimize its dimensions given the cost constraint.
Let's assume the length of each side of the square base is x feet, and the height of the box is h feet.
The cost of the bottom and top of the box can be calculated as the area of two squares:
2 * x^2 * $2/ft^2 = $4x^2
The cost of the four sides of the box can be calculated as the area of four rectangles:
4 * x * h * $1/ft^2 = $4xh
The total cost of the box is the sum of the costs of the bottom/top and the sides:
Total Cost = $4x^2 + $4xh
Given that the total cost should be $12, we can set up the following equation:
$12 = $4x^2 + $4xh
Simplifying the equation, we get:
3 = x^2 + xh
Now, we need to express h in terms of x to eliminate one variable. Let's rearrange the equation:
h = (3 - x^2)/x
To find the dimensions that maximize the value, we need to maximize the volume of the box, which is given by:
Volume = x^2 * h = x^2 * (3 - x^2)/x = 3x - x^3
Now, we can take the derivative of the volume function with respect to x to find the critical points:
dV/dx = 3 - 3x^2
Setting dV/dx equal to zero, we get:
3 - 3x^2 = 0
3x^2 = 3
x^2 = 1
x = ± 1
Since we are dealing with dimensions, we take the positive value for x:
x = 1
Now, to find the corresponding value of h, we substitute x = 1 into the equation h = (3 - x^2)/x:
h = (3 - 1^2)/1 = 2
Therefore, the dimensions of the box that maximizes its value for $12 are:
Length of each side of the base: 1 foot
Height of the box: 2 feet
To find the dimensions of the box with the greatest value, we need to consider the cost of materials for each part of the box.
Let's assume the length of one side of the square base is "x" feet.
The cost of the bottom and top of the box, each having an area of x^2 square feet, is:
2 * $2 * x^2 = $4x^2.
The cost of the four sides of the box, each having an area of x * h square feet, is:
4 * $1 * x * h = $4xh.
Given that the total cost for the box is $12, we can write the equation:
$4x^2 + $4xh = $12.
Dividing both sides of the equation by 4, we get:
x^2 + xh = 3.
Now, we want to maximize the value of the box, which is given by the volume, V = x^2h.
To solve for "h" in terms of "x", we can rewrite the equation above as:
h = 3/x - x.
Now, substitute this expression for "h" into the equation for the volume:
V = x^2(3/x - x).
Simplifying this equation further gives:
V = 3x - x^3.
To find the maximum value of V, we can take the derivative of V with respect to x and set it equal to zero:
dV/dx = 3 - 3x^2 = 0.
Solving for "x" gives us:
3x^2 = 3,
x^2 = 1,
x = ±1.
Since we are dealing with a physical dimension, we disregard the negative value and consider x = 1.
Therefore, the dimensions of the box with the greatest value that can be constructed for $12 are:
Length of one side of the square base = x = 1 foot,
Height of the box = h = 3/x - x = 3/1 - 1 = 3 - 1 = 2 feet.
So, the dimensions of the box are 1 foot by 1 foot by 2 feet.