The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field.

The Computer Laboratory

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.

The above links are provided as is. They are not affiliated with the Climb Foundation unless otherwise specified.

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).

Hints

  • 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?

Solution

\(y_1\) 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 \(y_1=-4x+16.\)

If you have queries or suggestions about the content on this page or the CSAT Practice Platform then you can write to us at oi.foo[email protected]. Please do not write to this address regarding general admissions or course queries.