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 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 poin... 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

7 (a) 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 need to identify the height and the base of the triangle.

Let the coordinates of point Q be $(q, 15 - q^2)$ and point P be $(q, 0)$ , since Q lies vertically above q on the x-axis. The height of the triangle is thus the difference in the y-coordinates:

$h = 15 - q^2$

The length of the base, OP, is equal to $2q$ (as it extends from $-q$ to $q$ on the x-axis). Therefore, the area of the triangle can be calculated using the formula for the area of a triangle:

$A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2q \times (15 - q^2)$

Simplifying:

$A = q(15 - q^2) = 15q - q^3$

Thus, we have shown that the area of triangle OPQ is given by:

$A = 15q - q^3$

where $c$ remains to be determined.

Step 2

7 (b) Find the exact maximum area of triangle OPQ.

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Answer

To find the maximum area of triangle OPQ, we need to differentiate the area function with respect to $q$ and set the derivative equal to zero to find critical points.

The area function is:

$A(q) = 15q - q^3$

Differentiating:

$\frac{dA}{dq} = 15 - 3q^2$

Setting the derivative equal to zero:

$15 - 3q^2 = 0$ $3q^2 = 15$ $q^2 = 5$ $q = \sqrt{5}$

To confirm this value yields a maximum, we can check the second derivative:

$\frac{d^2A}{dq^2} = -6q$

At $q = \sqrt{5}$ , the second derivative is negative:

$\frac{d^2A}{dq^2} = -6 \sqrt{5} < 0$

Thus, $q = \sqrt{5}$ is indeed a maximum. Now, substituting back into the area function to find the maximum area:

$A(\sqrt{5}) = 15(\sqrt{5}) - (\sqrt{5})^3 = 15\sqrt{5} - 5\sqrt{5} = 10\sqrt{5}$

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

7 (a) Show that the area, A, of the triangle OPQ is given by A = 15q - q^3

7 (b) Find the exact maximum area of triangle OPQ.

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