Q.Find two numbers whose sum is 24 and whose product is as large as possible. OR Find the equation of the tangent and normal to the curve $x^{2/3} + y^{2/3} = 2$ at $(1, 1)$.

Maximise $xy^3$ under $x+y=60$ using the first/second derivative test; the numbers are $x=15,\ y=45$.

Express the product in one variable using $x+y=60$, then maximise with the first derivative; the numbers are $x=15,\ y=45$.

Maximise $P=x(24-x)$ using calculus for Part 1; differentiate the curve implicitly and use point-slope form for Part 2 (OR).

Part 1: Let the two numbers be $x$ and $24-x$. Their product:

$P(x) = x(24-x) = 24x-x^2$

$P'(x) = 24-2x = 0 \Rightarrow x=12$

$P''(x) = -2 < 0 \Rightarrow \text{maximum at } x=12$

The two numbers are $12$ and $24-12=12$; maximum product $=12\times12=144$.

OR — Part 2: Curve $x^{2/3}+y^{2/3}=2$ at $(1,1)$. Differentiating implicitly:

$\dfrac23 x^{-1/3} + \dfrac23 y^{-1/3}\dfrac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx} = -\left(\dfrac{y}{x}\right)^{1/3}$ …

Writing the product as $P=x(24-x)$ and maximising gives $x=12$, so both numbers are $12$.

Concept. To maximise a quantity subject to a constraint, express it as a one-variable function, then use $\dfrac{dP}{dx}=0$ and the second-derivative test.

Step-by-step. Let the numbers be $x$ and $24-x$. Their product is

$P(x)=x(24-x)=24x-x^2.$

$P'(x)=24-2x.$

Set $P'(x)=0$: $24-2x=0\Rightarrow x=12$. …