SOLUTION: E. Solve the problem.
A rectangular box with volume 468 cubic feet is built with a square base and top. The cost is $1.50 per square foot for the top and the bottom and $2.00 p
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A rectangular box with volume 468 cubic feet is built with a square base and top. The cost is $1.50 per square foot for the top and the bottom and $2.00 p
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Question 145552: E. Solve the problem.
A rectangular box with volume 468 cubic feet is built with a square base and top. The cost is $1.50 per square foot for the top and the bottom and $2.00 per square foot for the sides. Let x represent the length of a side of the base in feet. Express the cost of the box as a function of x and then graph this function. From the graph find the value of x, to the nearest hundredth of a foot, which will minimize the cost of the box.
F.
If an object is dropped from a tower, then the velocity, V (in feet per second), of the object after t seconds can be obtained by multiplying t by 32 and adding 10 to the result. Find V as a linear function of t, and use this function to evaluate V(3.3), the velocity of the object at time t = 3.3 seconds.
You can put this solution on YOUR website! A rectangular box with volume 468 cubic feet is built with a square base and top. The cost is $1.50 per square foot for the top and the bottom and $2.00 per square foot for the sides. Let x represent the length of a side of the base in feet. Express the cost of the box as a function of x and then graph this function. From the graph find the value of x, to the nearest hundredth of a foot, which will minimize the cost of the box.
:
Let h = the height of the box
the area of the bottom = x^2
Therefore:
x^2*h = volume
x^2*h = 468
Find h
h =
:
Area of the sides = x*h
Substituting for h
Area of the sides = x() =
Cost of 4 sides = 2(4()) =
:
Cost of the bottom and the top = 1.50(2x^2) = 3x^2
:
Total cost = f(x)
:
f(x) = 3x^2 +
Use this equation to plot a graph y = f(x) = cost
:
I graphed it on my TI83, and found the minimum:
minimum cost at; x = 8.545 ft,
:
:
F.
If an object is dropped from a tower, then the velocity, V (in feet per second), of the object after t seconds can be obtained by multiplying t by 32 and adding 10 to the result. Find V as a linear function of t, and use this function to evaluate V(3.3), the velocity of the object at time t = 3.3 seconds.
:
The given equation: V(t) = 32t + 10
:
V(3.3) = 32(3.3) + 10
V(3.3) = 115.6 ft/sec after 3.3 sec