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Home > Maths > Calculus > Differentiation Part 2


Differentiation Part 2


Knowledge of stationary points and curve sketching provide a method of determining the greatest and least values of a function within a given interval.

Work your way through the following example.


A rectangle ABCD measures 7 units by 11 units, as shown.

Triangle AFE

Triangle AFE

E lies on BC x units from B

F lies on CD 2x units from C

Triangle AFE is shaded

  1. Show that the area, A units2, of the shaded region is given by A = x^2 - {{11} \over 2}x + {{77} \over 2}
  2. Find the value of x for which the shaded area is least.
  1. This part of the question just requires you to work out the area of the rectangle then subtract the areas of the three white triangles.

    \eqalign{ A & = & 7\times 11 - \;\; (\frac {1}{2}(11\times x) + \frac {1}{2}2x (7 - x) + \frac{1}{2}7 (11 - 2x)) \cr & = & 77 - {1 \over 2}\;\; (11x + 14x - 2x^2 + 77 - 14x) \cr & = & 77 - {1 \over 2}\;\;(77 + 11x - 2x^2 ) \cr & = & {{77} \over 2} - {{11} \over 2}x + x^2 \cr}

    This is the modelling part of the question, which is generally found to be more difficult than the actual optimisation. The question is deliberately framed so you can still score full marks in part 2, even if you can't do part 1. If, for part 1, you get a different formula from the one in the question, don't use that formula in part 2, use the formula given.

  2. Because this function is quadratic, we can solve the problem without calculus. In general, the extreme values are found by differentiation

    {{dA} \over {dx}} = 2x - {{11} \over 2} = 0 for stationary points \Rightarrow x = {{11} \over 4}

    We can justify the minimum from a nature table, but it is easier to note that this is a quadratic function with + 1x^2.

    The shaded area is least for x = \frac{11}{4}\; units.

    (Strictly speaking 0 < x ≤ 5.5, so we should also consider the end points when looking for extrema. We can omit that here because the graph is a parabola.)

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