Maximizing volume

theowne · #1 $Old$ April 30th, 2008, 10:39 PM

A box shaped block has a length equal to twice the width and the total surface area is 200cm^3. Find the dimensions of the maximum volume of the block:

This was my functions

Volume = [w(2w)] multiplied by (d)

SA = 2[w(2w)]+2(dw)+2(d(2w))

So

d = (200-4w^2)/(2(w+2w))

Is that right? How would you approach this question?

earboth · #2 $Old$ April 30th, 2008, 11:52 PM

Quote:

Originally Posted by theowne $View Post$

A box shaped block has a length equal to twice the width and the total surface area is 200cm^3. Find the dimensions of the maximum volume of the block:

This was my functions

Volume = [w(2w)] multiplied by (d)

SA = 2[w(2w)]+2(dw)+2(d(2w))

So

d = (200-4w^2)/(2(w+2w))

Is that right? How would you approach this question?

All your considerations and calculations are OK!

Take your last result and simplify it a little bit:

$d = \frac{200-4w^2}{2(w+2w)}=\frac{200-4w^2}{6w}=\frac{100}{3w} - \frac23 w^2$

Now plug in this term for d into the equation of the volume. You'll get the equation of a function calculating the volume with respect to w:

$V(w)=2w^2 \cdot \left(\frac{100}{3w} - \frac23 w^2 \right) = \frac{200}3 w - \frac43 w^4$

Now calculate the domain of the function, then the first derivative of V, solve for w the equation $V'(w) = 0$

I'll asumme that you can handle this procedure.