Question

solve using differentiation please.

a closed box, whose length is twice its width, is to have a surface 192 square inches. find the dimension of the box when the volume is maximum.
answer is 4x8x5 1/3 in.

solution please. :)

Answer 1

Find a formula for the volume of the box.
Use the surface area to express the unknown variable in terms of x.

Differentiate the volume with respect to x.
Set this to 0 to find the max value.

Answer 2

value of x when f ' (x) = 0 = maximum value

Answer 3

i still don't get the write answer. i have tried differentiating the volume with respect to x. still it's wrong.

Answer 4

The total surface area of a box of width W, length L and height H is the surface area of the six rectangular sides, therefore\[2(WL+LH+WH)=192.\]Additionally, it is given that L=2W. Putting these together and doing a little rearranging gives \[2W^2 +3WH = 96\]and solving for H gives:\[ H = (96-2W^2)/3W\]The volume of the box is WLH, so substituting in our equations for H and L in terms of W:\[Volume = (192W-4W^3)/3 = 64W - (4/3)W^3\]Differentiate V with respect to W to get:\[dV/dW = 64 - 4W^2\]As noted above, the maximum volume is when dV/dW is set to zero, so:\[64-4W^2=0\]Solving for W gives W is 4, and hence L is 8 and H is 5.33 by substituting back into the equations derived above.

Answer 5

you are so great! many thanks for the complete solution. :)

Answer 6

No problem - the same method should work with any similar question about volume/surface area maximisation too :)