# SQA Higher Maths 2010 Paper II Q5

 2010 Paper II Q 5 In the below x^2 means “x squared” or x times x etc, sqrt(x) means the “square root of x”, and I’m using * for the multiplication sign. Note that x can vary from a minimum of 0 but is bounded on the right by the upper parabola (the green parabola in my diagram) where the point Q lies. As x changes horizontally the height of the of the rectangle PQRS varies and therefore so does the area of the rectangle. In part (a) (i) we need to get an expression for the height and in (a) (ii) the area of the rectangle. The thing to see here is that the rectangular area PQRS is double the area of TPQM (see my diagram below for M). TP is x units. In order to get the area of TPQM we need to get an expression for the height of the rectangle ie QP. The point Q lies on the upper parabola hence its y co-ordinate is 10 - x^2. The point P lies directly below it on the line y = 4 hence its y co-ordinate is 4. The length QP is therefore the difference between these two y co-ordinates ie 10 - x^2 - 4 = 6 - x^2 Therefore the area of TPQM is x(6 - x^2) and therefore the area of PQRS is 2x(6 - x^2) = 12x - 2x^3 To find the value of x which gives the maximum area first differentiate A(x) with respect to x A’(x) = 12 - 6x^2 = 6(2 - x^2) NB this is a quadratic expression with a -ve x^2 term, so the parabola is a “sad” one i.e. one with a maximum turning point. Since we need to find a maximum area this is a good sign that our working so far has been correct. At a turning point the slope, or gradient, of the tangent to the curve is 0. Since the derivative is an expression which gives us the slope of the tangent line we need to find the value of x which makes the derivative 0. ie 6(2 - x^2) = 0 ie 2 - x^2 = 0 ie x = sqrt(2) Now use the expression for the area found at part (a) (ii) A(sqrt(2)) = 12 * sqrt(2) - 2 * sqrt(2)^3 = 12 * sqrt(2) - 4 * sqrt(2) — why so? The above is equal to 8 * sqrt(2)