SOLUTION: Suppose you want to make a closed box with a square base. The material for the top and the bottom of the box costs $2 per sq in and the material for the sides costs $3 per sq in. W
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-> SOLUTION: Suppose you want to make a closed box with a square base. The material for the top and the bottom of the box costs $2 per sq in and the material for the sides costs $3 per sq in. W
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Question 1000879: Suppose you want to make a closed box with a square base. The material for the top and the bottom of the box costs $2 per sq in and the material for the sides costs $3 per sq in. What are the dimensions of the box of the greatest volume that can be constructed for $60? Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! you can solve this graphically or you can solve it using calculus.
i don't know any other ways.
the graphical solution is shown below:
that 2.236 turns out to be sqrt(5).
the value of x is sqrt(5).
the value of y is 7.454
x is the length of a side of the square base.
y is the volume.
the formula for volume is derived as follows:
s = measure of one of the sides of the base.
h = measure of the height.
area of the base = s^2
volume = s^2*h
surface area is derived as follows:
sa = 2*s^2 + 4*h*s
cost of surface area is derived as follows:
cost of the base material is equal to 2 * the area of the base.
cost of the side material is equal to 3 * the area of the sides between the 2 bases.
the total cost is there equal to 2 * area of the bases plus 3 * area of the side faces.
you get total cost = 2 * (2s^2) + 3*(4hs) which becomes:
total cost = 4s^2 + 12hs
since total cost is 60, you get:
60 = 4s^2 + 12hs
in this equation, you can solve for h as follows:
subtract 4s^2 from both sides to get:
60 - 4s^2 = 12hs
divide both sides by 12s to get:
(60-4s^2) / 12s = h
the volume is equal to s^2*h
replace h with (60-4s^2)/12s to get:
volume = s^2 * (60-4s^2)/12s
factor out an s from the numerator and denominator and you get:
volume = s*(60-4s^2)/12
to graph this equation, make volume = y and make s = x.
formula becomes:
y = x*(60-4x^2)/12
this is the equation that was graphed above.
to solve this using calculus, get the derivative of the equation and set it equal to 0 and solve for x to find the maximum point on the graph.
the derivative turns out to be y' = 5-x^2
set it equal to 0 and you get 0 = 5-x^2
add x^2 to both sides to get x^2 = 5
take square root of both sides to get x = +/- sqrt(5).
it has to be sqrt(5) because negative values are not allowed.
the derivative is telling you that you need to evalute your equation at x = sqrt(5) to find the maximum volume.
that turns out to be same as what the graph is showing you.
the more detailed answer is:
x = s = 2.236068
y volume = 7.4535599
round these to x = 2.236 and y = 7.454
x is equal to s which is the width
y is the volume
the volume is s^2*h
you can use this formula to solve for h.
from this formula, solve for h to get h = v/s^2.
that becomes h = 7.4535599/2.236068 = 1.49071195
round to 3 decimal place to get h = 1.491
the dimension of the box with the greatest volume that has a surface area that costs 60 dollars is therefore:
s = 2.236
h = 1.491
if you recall, we made s = x when we graphed it.
when we found x, we automatically found s because they're equivalent to each other.
surface area = 2 * s^2 + 4hs
cost for surface area = 4s^2 + 12hs
when s = 2.236 and h = 1.491, we get cost for surface area = 59.999999999 which rounds to 60.
we got the maximum volume for a box that has a total cost for surface area of 60 dollars.
derivative was found using the following derivative calculator.
the calculator is very useful when you're not sure how to find the derivative.
