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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).

## 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 $$y_1=-4x+16.$$

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