1000cm^3 of metal is to be cast as a rectangular block with square ends (Calculus Optimization question)

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 )

• The formula $l = (1000-2x^2)/4x$ is incorrect. I think you used the formula for the area when you should have used a formula for volume. – David K Dec 26 '18 at 22:19

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.

• Thank you so much, an oversight on my part. The formatting in this answer is brilliant, well done! – Mahdi Shah Dec 27 '18 at 12:54
• A pleasure to have helped! – KM101 Dec 27 '18 at 13:04