Question

Question #fa51c

Answer 1

`256(sqrt(3))/9`

`49.267 cm^3` (3. d.p.)

Explanation:

I have omitted the calculations for finding the derivatives, as I assume at this level you know how to perform these, and also to save the answer from be excessively long.

From the given information we have:

Surface Area = 64 `cm^2`

We require maximum volume.

Let `S` be surface area and `V` be volume.
Let `a` be the length of each side of the base of the box, and let `h` be the height of the box.

Surface Area:

`S= a^2+4ah= 64`

Volume:

`V= a^2h`

We can not differentiate this, because we have 2 variables. We need to find `h` in terms of `a`. We can use the surface area equation, since this has to be constant ( ie. we need this to be `64cm^2` regardless of volume).

`S= a^2+4ah= 64=>h=(64-a^2)/(4a)`

And then volume is:

`V= a^2((64-a^2)/(4a))=(64a-a^3)/4`

We then differentiate this:

`(dV)/(da) ((64-a^2)/(4a))=16 - 3/4 a^2`

We know that maximum/minimum and points of inflection occur when the gradient is `0`. So:

`(dV)/(da) =0`

`16 - 3/4 a^2=0=> a = -8sqrt(3)/3 and 8sqrt(3)/3` ( negative value not applicable for length or volume ).

And:

`h = (64-a^2)/(4a)= (64-(8sqrt(3)/3)^2)/(4(8sqrt(3)/3))=(4sqrt(3))/3`

We know that for a maximum value that the second derivative is negative.

So:

`(d^2V)/(da^2)= -3/2a`

Plugging in `a`

`-3/2(8sqrt(3)/3)=-4sqrt(3)` (so this is a maximum value)

Our values for `a and h` are.

`color(blue)(a = 8sqrt(3)/3)`

`color(blue)(h = (4sqrt(3))/3)`

Maximum volume is:

`V=(8sqrt(3)/3)^2((4sqrt(3))/3)=color(blue)(256(sqrt(3))/9)` ,`49.267 cm^3` (3. d.p.)

We should check that the surface area is still `64cm^2`

`(8sqrt(3)/3)^2+4*(8sqrt(3)/3)*(4sqrt(3)/3)=64` (as required)

This is the graphs of the function and the first derivative, so you can see how maximum and minimum points of the function line up with the slope of the derivative. From this you will be able to see why we use the second derivatives negative or positive output to determine max/min values.