Optimising rectangle area between two parabolas

Key concept: This question combines geometric constraints with algebraic modelling to produce and maximize a cubic area function.

Step by step

Set up the model

The rectangle is placed between the parabolas

$y=16-4x^2 \quad \text{and} \quad y=\frac18(16-4x^2).$

The lower vertices lie on the x-axis, and the top vertices touch the upper parabola. The rectangle has length $x$ units, so its width is related to the horizontal position of the vertices and its height comes from the parabola value.

The problem states that the total area is

$A(x)=6x-x^3.$

That is the function to maximize.

Why the area is a cubic

The upper parabola gives the height of the rectangle, and the horizontal dimension contributes another factor of $x$. When the geometry is simplified, one linear factor and one quadratic factor combine to form a cubic polynomial. The given result

$A(x)=6x-x^3$

captures that relationship exactly.

Maximise the area

Differentiate:

$A'(x)=6-3x^2.$

Set the derivative equal to zero:

$6-3x^2=0$

$3x^2=6$

$x^2=2$

$x=\sqrt{2}$

since $x$ must be positive in this geometric context.

Now check that this gives a maximum. The second derivative is

$A''(x)=-6x,$

which is negative for positive $x$. So $x=\sqrt{2}$ gives a maximum area.

Substitute into the area function:

$A(\sqrt{2})=6\sqrt{2}-(\sqrt{2})^3=6\sqrt{2}-2\sqrt{2}=4\sqrt{2}.$

So the maximum area is

$4\sqrt{2}\text{ units}^2.$

Final answer

The value of $x$ that maximises the area is $\sqrt{2}$, and the maximum area is $4\sqrt{2}\text{ units}^2$.

Useful check

Because $A(x)=6x-x^3$ is an odd cubic with a positive linear term and negative cubic term, it rises first and then falls. That shape matches the existence of a single positive maximum.

Pitfall alert

A common mistake is to ignore the domain. Here, $x$ represents a physical length, so negative solutions from $x^2=2$ must be rejected. Another error is to stop at the derivative and forget to verify that the stationary point is a maximum. For cubic optimisation, the second derivative test or a sign chart is a good safeguard. Some students also mix up area and height, which leads to the wrong function entirely.

Try different conditions

If the area function were changed to $A(x)=8x-x^3$, the same derivative method would give $A'(x)=8-3x^2$, so the maximizing value would be $x=\sqrt{8/3}$. If the rectangle had to fit between a different pair of curves, the first job would be to rebuild the area formula from the new geometry before differentiating. The optimization method would stay the same even though the constants change.

Further reading

cubic function, stationary point, second derivative test