The volume V cm³ of a box, of height x cm, is given by $$ V = 4x(5-x)^2 $$ , 0 < x < 5 (a) Find \( \frac{dV}{dx} \) - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 3

To find the maximum volume, we need to set ( \frac{dV}{dx} = 0 ). From our earlier work:

$4(5-x)(5-3x) = 0$

This gives solutions:

1. ( 5-x = 0 \Rightarrow x = 5 ) (not in range)
2. ( 5-3x = 0 \Rightarrow x = \frac{5}{3} )

Next, we evaluate the volume at this critical point:

$V \left( \frac{5}{3} \right) = 4 \cdot \frac{5}{3} \cdot \left( 5 - \frac{5}{3} \right)^2$

Calculating further:

$= 4 \cdot \frac{5}{3} \cdot \left( \frac{10}{3} \right)^2 = 4 \cdot \frac{5}{3} \cdot \frac{100}{9} = \frac{2000}{27} \text{ cm}³$

Thus, the maximum volume is ( \frac{2000}{27} \text{ cm}³ ).

To confirm that ( \frac{5}{3} ) yields a maximum, we can use the second derivative test. We compute the second derivative:

$\frac{d^2V}{dx^2}$

From our first derivative:

$\frac{dV}{dx} = 4(5-x)(5-3x)$

Differentiate again:

Using the product rule we find:

$\frac{d^2V}{dx^2} = -12(5-x) + 12(5-3x) \Rightarrow \frac{d^2V}{dx^2} = -12(5-3x)$

Now evaluate at ( x = \frac{5}{3} ):

$\frac{d^2V}{dx^2} = -12 \cdot (5-5) = -0$

To ensure concavity, check values around ( \frac{5}{3} ):

For ( x < \frac{5}{3} ), ( \frac{dV}{dx} > 0 ) (increasing) and for ( x > \frac{5}{3} ), ( \frac{dV}{dx} < 0 ) (decreasing).

Thus, ( x = \frac{5}{3} ) is indeed a maximum.