Modelling with Differentiation and Optimisation Flashcards

All 4 cards in this deck

What are the steps to use calculus to find the maximum value of a quantity?
e.g. $V = 450x^2 - 30x^3$

1. Differentiate: $\frac{dV}{dx} = 900x - 90x^2$
2. Set $\frac{dV}{dx} = 0$ and solve: $x = 10$
3. Substitute back into $V$: maximum value $= 15\,000$

What are the steps to set up and optimise a quantity in a practical problem?
e.g. maximise the area of a rectangle of perimeter 20 cm

1. Write the quantity in terms of one variable: $A = x(10 - x)$
2. Differentiate and set to $0$: $10 - 2x = 0$, so $x = 5$
3. Substitute back: maximum area $= 25$ cm$^2$

What are the steps to show that a turning value of $V$ is a minimum?
e.g. $\frac{dV}{dx} = 3x^2 - 12$ with turning point at $x = 2$

1. Differentiate again: $\frac{d^2V}{dx^2} = 6x$
2. Substitute the turning point: $\frac{d^2V}{dx^2} = 12$
3. It is positive, so $V$ has a minimum.

True or false? The value of $x$ found from solving $\frac{dV}{dx} = 0$ is the maximum value of $V$.

False. That $x$ must be substituted back into the expression for $V$ to give the maximum value.