You are given a piece of sheet metal that is twice as long as it is wide and has an area of 800m^2. Find the dimensions of the rectangular box that would contain a maximum volume if it were constructed from this piece of metal by cutting squares of equal area at all four corners and folding up the sides. The Box will not have a lid.
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Let:
w be the width of the pen
L be the length of the pen
The area (A) is:
A = wL
Since each pen is 2400 m^2, then
wL = 2400
If we solve for w, we get:
w = 2400/L
If we create the pens so they share a common length, then we have 10 widths and 6 lengths to add together to determine the length of fencing (F), or:
F = 10w + 6L
Substituting our solution for w from above, we get:
F = 10*2400/L + 6L
F = 24000/L + 6L
Drawing the function out we can see that there is a minimum somewhere (if you go to the Wolfram link I gave you, it may tell you they need more time to calculate...accept that). To find the minimum we must find dF/dL and set dF/dL to zero and solve for L. To find dF/dL we differentiate F with respect to L we get:
dF/dL = -24000/L^2 + 6
Setting dF/dL to zero and solving for L:
dF/dL = 0 = -24000/L^2 + 6
6 = 24000/L^2
6L^2 = 24000
L^2 = 4000
L = ±√4000
L = ±20√10
Since L must be greater than 10m, we can ignore the negative value, so L = 20√10m.
To find w, we use our initial solution:
w = 2400/(20√10)
w = 120/√10
Multiplying by √10/√10 is the same as multiplying by 1:
w = 120√10/(√10*√10)
w = 120√10/10
w = 12√10m
The dimensions to minimize the amount of fencing used are 12√10m by 20√10m.
w be the width of the pen
L be the length of the pen
The area (A) is:
A = wL
Since each pen is 2400 m^2, then
wL = 2400
If we solve for w, we get:
w = 2400/L
If we create the pens so they share a common length, then we have 10 widths and 6 lengths to add together to determine the length of fencing (F), or:
F = 10w + 6L
Substituting our solution for w from above, we get:
F = 10*2400/L + 6L
F = 24000/L + 6L
Drawing the function out we can see that there is a minimum somewhere (if you go to the Wolfram link I gave you, it may tell you they need more time to calculate...accept that). To find the minimum we must find dF/dL and set dF/dL to zero and solve for L. To find dF/dL we differentiate F with respect to L we get:
dF/dL = -24000/L^2 + 6
Setting dF/dL to zero and solving for L:
dF/dL = 0 = -24000/L^2 + 6
6 = 24000/L^2
6L^2 = 24000
L^2 = 4000
L = ±√4000
L = ±20√10
Since L must be greater than 10m, we can ignore the negative value, so L = 20√10m.
To find w, we use our initial solution:
w = 2400/(20√10)
w = 120/√10
Multiplying by √10/√10 is the same as multiplying by 1:
w = 120√10/(√10*√10)
w = 120√10/10
w = 12√10m
The dimensions to minimize the amount of fencing used are 12√10m by 20√10m.