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Practice Paper 2 Question 10

A line \(y_1=ax+b\) is tangent to the curve \(y_2=12-x^2,\) with \(0<x<\sqrt{12}.\) Find the reals \(a,b\) such that the area delimited by \(y_1,\) the \(x\)-axis and the \(y\)-axis is minimized.

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Warm-up Questions

  1. Find the equation of the line that is tangent to the curve \(y=(x-1)^3\) at \(x=0.\)
  2. A rectangular cuboid (rectangular prism) has sides of length \(x,\) \(x+2\) and \(5-x.\) Find the value of \(x\) that maximizes its volume (make sure it gives a maximum).


  • Hint 1
    Try expressing the coordinates of the triangle vertices with respect to a single (and appropriate) variable.
  • Hint 2
    For example, try to express the intersection points of the line with the \(x\) and \(y\) axes with respect to the tangent point \(x\) coordinate, call it \(t\)?
  • Hint 3
    What is the area of the triangle in terms of those intersection points?
  • Hint 4
    What about in terms of \(t\)?
  • Hint 5
    How does one find the maximum of a function of a single variable?


\(y_1\) crosses the \(x\) and \(y\) axes at \(y=b\) and \(x=-\frac{b}{a}\) respectively, and so the area in question (right triangle) is \(A=-\frac{b^2}{2a}.\) However, both \(a\) and \(b\) are functions of the \(x\)-coordinate of the tangent point, which we'll denote with \(t,\) i.e. \(a=a(t)\) and \(b=b(t),\) and hence \(A=A(t).\) We must minimize the latter.

At the tangent point \(x=t\) we equate the gradients (derivatives), and the values of the line and of the curve. Hence \(a=-2t\) and \(y_1(t)=y_2(t),\) i.e. \(b=12+t^2.\) Writing \(\frac{dA}{dt} = 0\) we have: \(\frac{d}{dt} \frac{(12+t^2)^2}{4t} =\frac{2(12+t^2)(2t)}{4t}-\frac{(12+t^2)^2}{4t^2} = 0,\) which gives \(4t^2=12+t^2,\) thus \(t=2\) and \(y_1=-4x+16.\)

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