The cost of making a table partly varies as the area and partly varies as square of the length. lf the cost of a table 2m by 3m is $50000 and the cost of a table 1.5m by 4m is $64000. Find the cost of a table 2m by 4m.
Let's assume that the cost of the table is directly proportional to the area and the square of the length. Therefore, we can write the equation:
Cost = k * (Area) * (Length^2)
where k is the constant of proportionality.
From the given information, we have two equations:
1) Cost1 = k * (Area1) * (Length1^2)
$50000 = k * (2m * 3m) * (2m^2)
2) Cost2 = k * (Area2) * (Length2^2)
$64000 = k * (1.5m * 4m) * (4m^2)
Let's divide equation 2 by equation 1 to eliminate k:
Cost2/Cost1 = (Area2 * Length2^2) / (Area1 * Length1^2)
($64000/$50000) = ((1.5m * 4m) * (4m^2)) / ((2m * 3m) * (2m^2))
Simplifying this expression:
1.28 = (6m^3) / (6m^3)
So, the value of 1.28 holds true for the ratio of the costs of two tables. Let's assume the cost of the table 2m by 4m is C:
C = 1.28 * Cost1
C = 1.28 * $50000
C = $64000
Therefore, the cost of a table 2m by 4m is $64000.
To find the cost of a table that is 2m by 4m, we can use the given information about the cost and dimensions of two other tables.
Let's assume that the cost of making a table partly varies linearly with the area and partly varies quadratically with the length. We can express this relationship using the equation:
Cost = k * Area + a * Length^2
where:
- Cost is the total cost of making the table
- Area is the area of the table (length multiplied by width)
- Length is the longer side of the table
- k and a are constants
We can now set up two equations based on the given information:
For the table that is 2m by 3m:
50000 = k * (2 * 3) + a * 3^2
= 6k + 9a
For the table that is 1.5m by 4m:
64000 = k * (1.5 * 4) + a * 4^2
= 6k + 16a
We now have a system of equations:
6k + 9a = 50000 (Equation 1)
6k + 16a = 64000 (Equation 2)
To find the values of k and a, we can subtract Equation 1 from Equation 2:
(6k + 16a) - (6k + 9a) = 64000 - 50000
7a = 14000
a = 14000 / 7
a = 2000
Substituting the value of a back into Equation 1, we can solve for k:
6k + 9(2000) = 50000
6k + 18000 = 50000
6k = 32000
k = 32000 / 6
k = 5333.33 (approx.)
Finally, we can use the obtained values for k and a in our original equation to find the cost of the table that is 2m by 4m:
Cost = k * Area + a * Length^2
= 5333.33 * (2 * 4) + 2000 * 4^2
= 5333.33 * 8 + 2000 * 16
= 42666.64 + 32000
= $74666.64
Therefore, the cost of a table that is 2m by 4m is approximately $74666.64.