Let's denote the following variables:

• A: Cost of material for the base per square unit (in dollars per square unit)
• B: Cost of material for the sides per square unit (in dollars per square unit)
• C: Cost of each bracket (in dollars)
• D: Cost to affix each bracket (in dollars)
• x: Length of the sides of the base (in units)
• h: Height of the box (in units)
• V: Volume specified by the Army (in cubic units)

Now, let's write an expression for the total cost of producing a box:

1. Cost of material for the base: The area of the base is x * x, and the cost is A per square unit. So, the cost for the base material is A * x^2.

2. Cost of material for the sides: The area of each side is 4 * x * h (there are four sides), and the cost is B per square unit. So, the cost for the side material is 4 * B * x * h.

3. Cost of brackets: There are 8 brackets, and each costs C. So, the cost for the brackets is 8 * C.

4. Cost to affix brackets: There are 8 brackets, and each costs D to affix. So, the cost to affix brackets is 8 * D.

Now, the total cost (TC) can be expressed as the sum of these costs:

$��=�\cdot {�}^2+4\cdot �\cdot �\cdot ℎ+8\cdot �+8\cdot �$

Now, Metal Containers wants to minimize the total cost while producing a box with a specified volume V. The volume of the box is given by:

$�={�}^2\cdot ℎ$

To minimize the cost with the volume constraint, you can use the volume formula to express one of the variables in terms of the other and substitute it into the cost function. In this case, you can express $ℎ$ in terms of $�$ using the volume equation:

$ℎ=\frac{�}{{�}^2}$

Now, substitute this expression for $ℎ$ into the total cost equation:

$��=�\cdot {�}^2+4\cdot �\cdot �\cdot \left(\frac{�}{{�}^2}\right)+8\cdot �+8\cdot �$

Given values:

• A = $12 per square foot (base material cost)
• B = $8 per square foot (side material cost)
• C = $5 per bracket
• D = $1 per bracket (labor cost per bracket)
• $�=48$ cubic feet (specified volume)

We know that $�={�}^2\cdot ℎ$, and the specified volume is $48$ cubic feet. To find the dimensions that minimize the cost, we need to express $ℎ$ in terms of $�$:

$�={�}^2\cdot ℎ$ $48={�}^2\cdot ℎ$

Solving for $ℎ$:

$ℎ=\frac{�}{{�}^2}$ $ℎ=\frac{48}{{�}^2}$

Now, substitute this expression for $ℎ$ into the total cost equation:

$��=12\cdot {�}^2+4\cdot 8\cdot �\cdot \left(\frac{48}{{�}^2}\right)+8\cdot 5+8\cdot 1$

Simplify the equation:

$��=12{�}^2+\frac{192}{�}+40+8$

Now, we want to minimize this cost. Take the derivative of $��$ with respect to $�$ and set it equal to zero:

$\frac{���}{��}=24�-\frac{192}{{�}^2}=0$

Multiply through by ${�}^2$:

$24{�}^3-192=0$

Solving for $�$:

${�}^3=\frac{192}{24}$ ${�}^3=8$

$�=2$

Now that we have the value of $�$, substitute it back into the expression for $ℎ$:

$ℎ=\frac{48}{2^2}$ $ℎ=12$

So, the dimensions that minimize the cost are $�=2$ feet and $ℎ=12$ feet.

Now, substitute these values into the total cost equation to find the minimum cost:

$��=12\cdot (2{)}^2+4\cdot 8\cdot 2\cdot \left(\frac{48}{(2{)}^2}\right)+8\cdot 5+8\cdot 1$

$��=48+192+40+8$

$��=288$

Therefore, the minimum cost is $288, and the dimensions of the box that meet the army's requirements at this minimum cost are $�=2$ feet and $ℎ=12$ feet.

:) HAPPY LEARNING

Now, we want to minimize this cost. Take the derivative of $��$ with respect to $�$ and set it equal to zero:

$\frac{���}{��}=24�-\frac{192}{{�}^2}=0$

Multiply through by ${�}^2$:

$24{�}^3-192=0$

Solving for $�$:

${�}^3=\frac{192}{24}$ ${�}^3=8$

$�=2$

Now that we have the value of $�$, substitute it back into the expression for $ℎ$:

$ℎ=\frac{48}{2^2}$ $ℎ=12$

So, the dimensions that minimize the cost are $�=2$ feet and $ℎ=12$ feet.

Now, substitute these values into the total cost equation to find the minimum cost:

$��=12\cdot (2{)}^2+4\cdot 8\cdot 2\cdot \left(\frac{48}{(2{)}^2}\right)+8\cdot 5+8\cdot 1$

$��=48+192+40+8$

$��=288$

Therefore, the minimum cost is $288, and the dimensions of the box that meet the army's requirements at this minimum cost are $�=2$ feet and $ℎ=12$ feet.