the surface area cost of construction per m^2 for a large box is $50 for the base, $60 for the top and $40 for the walls. Find the dimensions of this box that minimize the cost if its volume has to be 9m^3 and its height, 1m

let the base be x by y m

V = (x)(y)(1) = xy
xy = 9
y = 9/x

cost = 50xy + 60xy + 2(40x) + 2(40y)
= 110xy + 80x + 80y
= 110x(9/x) + 80x + 720/x
= 110 + 80x + 720/x

d(cost)/dx = 80 - 720/x^2
= 0 for a min of cost
80x^2 = 720
x^2 = 9
x = 3
then y = 9/3 = 3

the box should be 3 by 3 by 1 m

To find the dimensions of the box that minimize the cost, we can set up an optimization problem and use calculus to solve it.

Let's start by finding the expressions for the cost and the surface area of the box:

Cost = (Cost of the base) + (Cost of the top) + (Cost of the walls)
Surface Area = (Area of the base) + (Area of the top) + (Area of the walls)

Given that:
Cost of the base = $50/m²
Cost of the top = $60/m²
Cost of the walls = $40/m²
Height of the box (h) = 1m
Volume of the box = 9m³

To find the area of the base:
Area of the base = length × width
Since the box has a fixed volume of 9m³, we know that the volume of the base must also be 9m³:
length × width × height = 9m³
Since the height is given as 1m, we can substitute h = 1 m into the equation:
length × width × 1m = 9m³
length × width = 9m²

Now we have expressions for the area of the base, the area of the top, and the area of the walls. Let's substitute these expressions into the equation for the surface area:

Surface Area = (Area of the base) + (Area of the top) + (Area of the walls)
Surface Area = (length × width) + (length × width) + (2 × length × height + 2 × width × height)
Surface Area = 2(length × width) + 2(width + length) × height

Now, we can express the cost in terms of the surface area:

Cost = (Cost of the base) × (Area of the base) + (Cost of the top) × (Area of the top) + (Cost of the walls) × (Area of the walls)
Cost = $50/m² × (length × width) + $60/m² × (length × width) + $40/m² × (2(length + width) × height)
Cost = ($50 + $60) × (length × width) + $40/m² × 2(length + width) × height
Cost = $110 × (length × width) + $80/m² × (length + width) × height

To minimize the cost, we can differentiate the cost expression with respect to either length or width and find its critical points. Let's differentiate with respect to length:

dCost/dlength = 110 × width + 80/h × (length + width)

To find the critical point, set the derivative equal to 0:

110 × width + 80/h × (length + width) = 0

Now, let's differentiate dCost/dlength again with respect to width to determine if this point is a minimum or maximum:

d²Cost/dwidth² = 110 + 80/h

Since the second derivative is positive (110 + 80/h > 0), we can conclude that the critical point is a minimum.

Now, let's find the dimensions that minimize the cost by solving the system of equations:

110 × width + 80/h × (length + width) = 0 (Equation 1)
length × width = 9m² (Equation 2)

We have two equations and two unknowns (length and width). We can solve this system of equations to find the values of length and width.

Once we have the values of length and width, we can use Equation 2 (length × width = 9m²) to find the specific dimensions of the box that minimize the cost.