The curve $y = 15 - x^2$ and the isosceles triangle OPQ are shown on the diagram below. Vertices P and Q lie on the curve such that Q lies vertically above some point $(q, 0)$. The line PQ is parallel to the x-axis. 7 (a) Show that the area, $A$, of the triangle OPQ is given by $A = 15q - q^3$ where $c$ is a constant to be found. 7 (b) Find the exact maximum area of triangle OPQ. Fully justify your answer.

A-Level AQA Maths Pure Applications of Differentiation: The curve $y = 15

The curve $y = 15 - x^2$ and the isosceles triangle OPQ are shown on the diagram below - AQA - A-Level Maths Pure - Question 7 - 2022 - Paper 2

Question 7

The curve $y = 15 - x^2$ and the isosceles triangle OPQ are shown on the diagram below. Vertices P and Q lie on the curve such that Q lies vertically above some poi... show full transcript

Worked Solution & Example Answer:The curve $y = 15 - x^2$ and the isosceles triangle OPQ are shown on the diagram below - AQA - A-Level Maths Pure - Question 7 - 2022 - Paper 2

Step 1

Show that the area, $A$, of the triangle OPQ is given by $A = 15q - q^3$

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Answer

To find the area of triangle OPQ, we first identify the height and base. The height of the triangle is the distance from point O (0, 0) to line PQ, which is defined by the equation of the curve:

The equation of the curve is given by: $y = 15 - q^2$
The base of the triangle, denoted as $q$ , is the horizontal distance from O to Q along the x-axis.
The area $A$ of triangle OPQ can be calculated using the formula for the area of a triangle: $A = rac{1}{2} \times \text{base} \times \text{height}$
Plugging in our expressions: $A = \frac{1}{2} \times (2q) \times (15 - q^2)$
This simplifies to: $A = q(15 - q^2)$ $A = 15q - q^3$
Therefore, we have shown that: $A = 15q - q^3$ where $c$ is a constant to be determined from the context.

Step 2

Find the exact maximum area of triangle OPQ.

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Answer

To find the maximum area of triangle OPQ, we need to optimize the area function $A = 15q - q^3$ .

First, we take the derivative of $A$ with respect to $q$ : $\frac{dA}{dq} = 15 - 3q^2$
We set the derivative equal to zero to find critical points: $15 - 3q^2 = 0$ $3q^2 = 15$ $q^2 = 5$ $q = \sqrt{5}$
Next, we check that this critical point is a maximum by examining the second derivative: $\frac{d^2A}{dq^2} = -6q$ Since $q = \sqrt{5}$ is positive, we find: $\frac{d^2A}{dq^2} < 0$ Thus, $q = \sqrt{5}$ corresponds to a local maximum.
We substitute $q = \sqrt{5}$ into the area function to find the maximum area: $A = 15(\sqrt{5}) - (\sqrt{5})^3$ $A = 15\sqrt{5} - 5\sqrt{5} = 10\sqrt{5}$
Therefore, the exact maximum area of triangle OPQ is: $A = 10\sqrt{5}$

The curve $y = 15 - x^2$ and the isosceles triangle OPQ are shown on the diagram below - AQA - A-Level Maths Pure - Question 7 - 2022 - Paper 2

Worked Solution & Example Answer:The curve $y = 15 - x^2$ and the isosceles triangle OPQ are shown on the diagram below - AQA - A-Level Maths Pure - Question 7 - 2022 - Paper 2

Show that the area, $A$, of the triangle OPQ is given by $A = 15q - q^3$

Find the exact maximum area of triangle OPQ.

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