SOLUTION: Hey guys, this question is related to applications of calculus (I couldn't find the section.
So, the question is as follows:
A rectangular block, the length of whose base is tw
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-> SOLUTION: Hey guys, this question is related to applications of calculus (I couldn't find the section.
So, the question is as follows:
A rectangular block, the length of whose base is tw
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Question 706282: Hey guys, this question is related to applications of calculus (I couldn't find the section.
So, the question is as follows:
A rectangular block, the length of whose base is twice the width has a total surface area of 300cm^2. Find the dimensions of the block if it is of maximum volume.
Thank You Found 2 solutions by josgarithmetic, ankor@dixie-net.com:Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website! x length, y width, z height.
volume of the box would be xyz.
x=2y as given, so volume is 2*y*y*z or volume .
Surface area was given as 300 cubic units. Accounting for the lengths measurements,
4y^2+2yz+4yz=300
...and few more steps...
z=(300-4y^2)/(2y+4y)
Back to the volume formula,
v=2y^2(2/3)(75-y^2)/y
v=... ----Taking derivative of formula in this form may be easiest.
The task to do now is the maximize v as a function of width y. Differentiate v with respect to y.
You then simply calculate x from knowing how x and y were given. Then you may need to use the z formula found above to caclulate z.
You can put this solution on YOUR website! A rectangular block, the length of whose base is twice the width has a total surface area of 300cm^2.
Find the dimensions of the block if it is of maximum volume.
:
Max volume will be when the larger dimension is the square sides
Let the squared base = 2w by 2w, the third dimension = w
:
Surface area then
2(2w*2w) + 4(2w*w) = 300
8w^2 + 8w^2 = 300
16w^2 = 300
w^2 = 300/16
w^2 = 18.75
w =
w ~ 4.33 cm is width and 8.66 cm the sides of the square base
:
Dimensions for max volume 8.66 by 8.66 by 4.33
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