Bill Ding plans to build a new hardware store. He buys a rectangular lot that is 50 ft by 200 ft, the 50-foot dimension being along the street. The store is to have an area of 4000 square feet. Construction costs $100 per linear foot for the part of the store along the street and only $80 per linear foot for the parts along the sides and back. To what dimensions should Bill build the store in order to minimize the construction coast?
Use the entire 200- ft dimension to
minimize cost:
200W = 4000 SQ FT,
W = 4000 / 200 = 20 FT.
L * W = 200 * 20.
To minimize the construction cost, we need to find the dimensions of the store that would result in the lowest total cost. Let's break this down step by step:
Step 1: Determine the dimensions of the store
Let's assume the store's length along the street is x feet (which is the dimension we need to find). The width of the store, which is perpendicular to the street, will then be (200 - x) feet.
Step 2: Calculate the area of the store
We know that the area of the store should be 4000 square feet. Therefore, we have:
Area = Length x Width
4000 = x * (200 - x)
Step 3: Simplify the equation
Simplifying the equation we obtained in step 2, we have:
4000 = 200x - x^2
Step 4: Rearrange the equation
Rearranging the equation to get it into a more usable form:
x^2 - 200x + 4000 = 0
Step 5: Solve for x using quadratic formula
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a = 1, b = -200, and c = 4000, we can find the value(s) of x.
Applying the quadratic formula, we have:
x = (-(-200) ± √((-200)^2 - 4(1)(4000))) / (2(1))
x = (200 ± √(40000 - 16000)) / 2
x = (200 ± √(24000)) / 2
x = (200 ± √(400 * 60)) / 2
x = (200 ± 20√(60)) / 2
x = 100 ± 10√(60)
Step 6: Determine the valid dimension
Since the length along the street (x) cannot be negative, we discard the negative value obtained in step 5. Therefore, we have:
x = 100 + 10√(60)
Step 7: Calculate the construction cost
Now that we have the value of x, we can determine the total construction cost.
The cost of the part along the street (50 ft) is $100 per linear foot, and the cost of the parts along the sides and back (150 ft) is $80 per linear foot.
Cost = (Cost per foot along street x Length along street) + (Cost per foot along sides and back x Total length along sides and back)
Cost = ($100 x 50) + ($80 x 150)
Cost = $5000 + $12000
Cost = $17000
So, to minimize construction costs, Bill Ding should build the store with dimensions of 100 + 10√(60) feet by 200 - (100 + 10√(60)) feet.
To find the dimensions that will minimize the construction cost, we need to formulate an equation for the cost based on the dimensions of the store.
Let's assume the width of the store (along the 200-foot side) is 'x' feet. Then the length of the store (along the 50-foot side) would be (4000/x) feet since the area of the store is 4000 square feet.
The cost of the store is calculated as follows:
- The two 50-foot walls (the ones along the street) cost $100 per linear foot. So, the cost for those two walls would be 2 * 50 * $100 = $10,000.
- The two x-foot walls (the ones along the sides) cost $80 per linear foot. So, the cost for those two walls would be 2 * x * $80 = $160x.
- The 200-foot wall (the one along the back) also costs $80 per linear foot. So, the cost for that wall would be 200 * $80 = $16,000.
Therefore, the total cost function (C) in terms of 'x' is given by: C(x) = 10,000 + 160x + 16,000.
To minimize the cost, we need to find the value of 'x' that minimizes this cost function.
To find the minimum of a quadratic function, we can take the derivative of the function with respect to 'x' and set it equal to zero. Then solve for 'x'.
Taking the derivative of C(x) with respect to 'x', we have:
C'(x) = 160
Setting C'(x) equal to zero, we get:
160 = 0
Since there are no solutions for this equation, that means there is no minimum or maximum value for 'x'. However, we can identify the endpoints of the feasible range.
The width, 'x', of the store must be within the range 0 < x < 200 feet.
Since it would not make sense to have a width of 0 feet or exceed the length of the lot (200 feet), we can conclude that 'x' should be between 0 and 200 feet.
In other words, Bill should build the store with a width that is greater than 0 feet and less than 200 feet. The length of the store would then be (4000/x) feet to maintain an area of 4000 square feet.