1000cm^3 of metal is to be cast as a rectangular block with square ends. Use calculus to show that for the least surface area a rectangular the rectangular block needs to be a cube.
Having some trouble, just learned about optimization. Really would appreciate help!
My working was:
2x^2 + 4xl = SA
dSA/dx = 4x + 4l
4x + 4l = 0 (to find maximum)
4x = -4l
x = -l
I just carried on knowing I was on the wrong track, would love corrections. :)
(Some more working was l = (1000-2x^2)/4x but if I substitute that I need to deal with imaginary numbers, which isn't in this course but is in Mathematics: Specialist, that's how I know about it. :P )
You have a rectangular prism with square ends, so the following can be said regarding volume
$$V_{block} = s^2l \implies 1000 = s^2l \tag{1}$$
where $s$ represents the length of the square sides.
For surface area, you have
$$A_{surface} = 2s^2+4sl \tag{2}$$
You’ve done everything correctly until this point.
However, there are two variables, so rewrite $(1)$ in terms of $l$, which becomes
$$l = \frac{1000}{s^2}$$
Plugging that in $(2)$, you get
$$A_{surface} = 2s^2+4s\left(\frac{1000}{s^2}\right)$$
$$A_{surface} = 2s^2+4000s^{-1}$$
Treat this as a function and take the derivative, yielding
$$\frac{dA}{ds} = 4s-4000s^{-2}$$
$$\frac{dA}{ds} = 0 \implies 0 = 4s-4000s^{-2} \implies s = 10$$
Plugging it into $(1)$, it becomes clear that $l = 10$ as well. Hence, the block must be a cube.