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Fake gold bricks are made by covering concrete cubes with gold paint, so the co...
Mar 29, 2024
Fake gold bricks are made by covering concrete cubes with gold paint, so the cost of the paint is proportional to their surface area while the cost of the concrete is proportional to their volume. If a 1 inch cube costs 1.30tomakewhilea2inchcubecosts1.30 to make while a 2 inch cube costs 6.80, then how much would a 3 inch cube cost?
Solution by Steps
step 1
To find the cost of a 3-inch cube, we first need to understand the relationship between the cost, surface area, and volume for the given cubes
step 2
Let's denote the cost to make a cube with side length s s inches as C(s) C(s) , the surface area of the cube as A(s)=6s2 A(s) = 6s^2 , and the volume of the cube as V(s)=s3 V(s) = s^3
step 3
The cost for a 1-inch cube is C(1)=$1.30 C(1) = \$1.30 , and for a 2-inch cube is C(2)=$6.80 C(2) = \$6.80
step 4
We can express the cost as C(s)=k1A(s)+k2V(s) C(s) = k_1A(s) + k_2V(s) , where k1 k_1 is the cost per unit surface area for the paint, and k2 k_2 is the cost per unit volume for the concrete
step 5
Using the costs for 1-inch and 2-inch cubes, we can set up a system of equations to solve for k1 k_1 and k2 k_2
step 6
Substituting A(1)=6 A(1) = 6 and V(1)=1 V(1) = 1 into C(1)=k1A(1)+k2V(1) C(1) = k_1A(1) + k_2V(1) , we get 1.3=6k1+k2 1.3 = 6k_1 + k_2
step 7
Substituting A(2)=24 A(2) = 24 and V(2)=8 V(2) = 8 into C(2)=k1A(2)+k2V(2) C(2) = k_1A(2) + k_2V(2) , we get 6.8=24k1+8k2 6.8 = 24k_1 + 8k_2
step 8
Solving the system of equations from steps 6 and 7, we find the values of k1 k_1 and k2 k_2
step 9
Once we have k1 k_1 and k2 k_2 , we can calculate the cost for a 3-inch cube using C(3)=k1A(3)+k2V(3) C(3) = k_1A(3) + k_2V(3) , where A(3)=54 A(3) = 54 and V(3)=27 V(3) = 27
step 10
Substitute k1 k_1 and k2 k_2 into the equation for C(3) C(3) to find the final cost
Answer
[Insert final answer here]
Key Concept
Cost Proportionality to Surface Area and Volume
Explanation
The cost of making the cubes is proportional to their surface area and volume, which means we can express the cost as a linear combination of the surface area and volume with constant coefficients. By solving the system of equations derived from the known costs of smaller cubes, we can find these coefficients and use them to calculate the cost of a larger cube.
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