joe is building a wooden cylindrical storage box. the box will have a volume og 6pi m^3. for design purposes, the thick wood for the top and the bottom costs $6/m^2 and the thinner wood for the sides cost $5/m^2.
a)express the cost of the wood as a function of the radius. make sure to simplify the expression.
b) if can afford to spend at most $72pi (~ $226.19) to make the storage box, what are the dimensions of the box?
cost=6*2PIr^2+5*2PIr*h
volume=PIr^2*h or h=volume/PIr^2
cost=12PIr^2+20PIr*6PI/PIr^2
can you please explain how you got these equations?
this is what i got:
cost=6*2PIr^2+5*2PIr*h
= 12PIr^2 + 10PIr*h
= 12PIr^2 + 10PIr*(6/r^2)
= 12PIr^2 + (60PI)/r
a) To express the cost of the wood as a function of the radius, let's start by considering the dimensions of the cylindrical storage box.
The formula to calculate the volume of a cylinder is:
V = πr^2h
where V represents the volume, r is the radius, and h is the height of the cylinder.
In this case, the volume of the box is given as 6π m^3. We can set up the equation as follows:
6π = πr^2h
Now, we need to solve this equation for h (height) in terms of r (radius). We divide both sides of the equation by πr^2:
6 = rh
Now, let's consider the surface area that needs to be covered by the wood. The top and bottom of the box are circles, and the sides form a rectangle when rolled out flat.
The formula for the surface area of a circle is:
A_circle = 2πr^2
The top and bottom combined will have a surface area of:
A_top_bottom = 2A_circle = 4πr^2
The formula for the surface area of a rectangle is:
A_rectangle = 2πrh
Adding the surface areas of the top and bottom to the sides, we get the total surface area:
A_total = A_top_bottom + A_rectangle = 4πr^2 + 2πrh
Now, we can determine the cost of the wood. The cost depends on the surface area covered.
For the top and bottom wood (thick wood), the cost per unit area is $6/m^2. So, the cost for the top and bottom combined is:
Cost_top_bottom = 4πr^2 * $6/m^2
For the side wood (thin wood), the cost per unit area is $5/m^2. So, the cost for the sides is:
Cost_sides = 2πrh * $5/m^2
To get the total cost, we add the costs of the top and bottom to the sides:
Cost_total = Cost_top_bottom + Cost_sides = 4πr^2 * $6/m^2 + 2πrh * $5/m^2
Simplifying the expression:
Cost_total = 24πr^2 + 10πrh
b) Now, let's determine the dimensions of the box using the budget of $72π (~ $226.19).
We have the cost of the wood as a function of the radius:
Cost_total = 24πr^2 + 10πrh
Let's set this expression equal to the maximum amount Joe can spend:
$72π = 24πr^2 + 10πrh
Divide both sides by π:
$72 = 24r^2 + 10rh
Simplify the equation further:
3 = r^2 + (10h/24)
Since we have two variables, we need another equation to solve for r and h simultaneously.
Remember the equation we derived from the volume of the cylinder:
6 = rh
Now, we can substitute this equation into our cost equation, replacing rh with 6:
3 = r^2 + (10*6/24)
Simplify:
3 = r^2 + (5/2)
Rearrange the equation:
r^2 = 3 - (5/2)
r^2 = 1/2
Taking the square root of both sides, we get:
r = ±√(1/2)
Since we're working with dimensions, we'll take the positive square root to have a positive radius value:
r = √(1/2)
Now, substitute this value back into the equation for finding h using the volume:
6 = √(1/2) * h
Solve for h:
h = 6 / √(1/2)
At this point, you can use a calculator to evaluate √(1/2) and h.
So, the dimensions of the box are: radius (r) = √(1/2) and height (h) = 6 / √(1/2).