a large closed storage rectangular box with a base constructed with two different types of wood. the base is made up of wood $5 per square feet and the top and sides are made of wood $3 per square feet suppose the amount available to spend is $1000 find the dimensions of the box with greatest volume
we want to maximize xyz subject to the constraint that
5xy+3(xy+2xz+2yz) = 1000
Using Lagrange multipliers, we find that x=y, and λ=2.1516 and z=4λ
That means a local max occurs at
(6.455,6.455,8.607)
A nice article on Lagrange multipliers, with an example similar to this problem, is at
http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx
To find the dimensions of the box with the greatest volume, we need to consider the cost and constraints. Let's assume the length of the box is L, the width is W, and the height is H.
We know that the base is made of wood that costs $5 per square foot, and the top and sides are made of wood that costs $3 per square foot. The total cost of the wood is limited to $1000.
First, let's calculate the cost of the base (which has two sides, each with an area of LW) and the cost of the top and sides (which have an area of 2LH + 2WH):
Cost of the base = 2LW * $5 = 10LW
Cost of the top and sides = (2LH + 2WH) * $3 = 6LH + 6WH
The total cost must be less than or equal to $1000, so we can write the equation:
10LW + 6LH + 6WH ≤ 1000
Now, let's consider the volume of the box. The volume of a rectangular box is given by V = LWH.
We want to find the dimensions that maximize the volume. This can be solved by taking the derivative of the volume function with respect to L, W, and H and equating them to zero. However, this process can be complex and may require calculus.
To simplify the problem, we can use a technique called optimization. Since we have an equation for the total cost, we can solve it for one variable and substitute it into the equation for the volume. This will give us a cost in terms of two variables, and we can focus on maximizing the volume.
Let's solve the cost equation for L:
10LW + 6LH + 6WH ≤ 1000
10LW ≤ 1000 - 6LH - 6WH
LW ≤ (1000 - 6LH - 6WH) / 10
L ≤ (1000 - 6LH - 6WH) / (10W)
Now, substitute this expression for L into the equation for volume:
V = LWH
V = [(1000 - 6LH - 6WH) / (10W)] * W * H
V = (1000WH - 6LH^2 - 6WH^2) / 10
Now, we have the volume equation in terms of W and H. To maximize the volume, we need to find the values of W and H that maximize the equation.
There are multiple ways to solve this equation, such as graphical methods or using optimization techniques. However, given the complexity of the equations, it is best to use optimization software or consult with a mathematician or engineer specialized in optimization problems to find the dimensions that maximize the volume.