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How can i find the point in the first quadrant on the parabola $$ y = 4-x ^ 2 $$ such that the triangle tangent to the parabola at the point and the coordinate axes has minimum area.

Some help to interpret the equation so minimize in this exercise i stuck i know how to calculate minima but i can't fin the equation to minimize.

I tried but $y = 4 - x^2$ and the tangent is the derivative then i have $dy/dx =-2x$ and i stuck here.Thanx for you hints and help.

Rosa Maria Gtz.
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  • You might want to rephrase the question to say, "...such that the right triangle in the first quadrant formed by the coordinate axes and the tangent line to the parabola" or something similar. – heropup Jun 06 '14 at 18:05

3 Answers3

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So you calculated the derivative, which means you know the slope of the tangent line for some general $x$-value. For a given point $(x_0, y_0)$ on this parabola (so in particular, $y_0 = 4-x_0^2$), what is the equation of the tangent line to this point?

Next, given the equation of the tangent line, where does it intersect the $x$- and $y$-axes? This will give you the length of the two legs of the right triangle enclosed by the tangent line.

Finally, what is the area of such a triangle as a function of the $x$-coordinate of the point $(x_0, y_0)$?

How can you minimize this area?

heropup
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ANSWER;

At the point (a,b), using the equation of the parabola

$$ b = 4-a^2$$

The equation of the tangent line is then

$$ y-b = -2a(x-a)$$

$$ y = b -2ax +2a^2$$

Substituting the value of b from the first equation,

$$ y = 4-a^2 - 2ax +2a^2 = 4 + a^2 - 2ax$$

For the point in X-axis,say $(x_1,0)$, $$ 0 = 4+a^2 - 2ax_1$$

$$ x_1 = \frac{4+a^2}{2a}$$ For the point in Y-axis, say $(0,y_1)$

$$ y_1 = 4+a^2$$

Area of the triangle $$= \frac{1}{2}x_1y_1$$

Substituting the value of $x_1,y_1$ in terms of a

$$ A = \frac{(4+a^2)^2}{4a}$$

Now set $\frac{dA}{da} = 0$

If you take the derivative using quotient rule, you get $$a = \frac{2}{\sqrt{3}}$$ and $$ b = \frac{8}{3}$$

Satish Ramanathan
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Using similar triangles, it can be proven that the area of the triangle is twice the product $xy$ where $y=f(x)$.

Maximizing $xy$ will give you your answer.

$A=xy=x(4-x^2)=4x-x^3$

$A'=0=4-3x^2$

So $x=\frac{2}{\sqrt{3}}$

TurlocTheRed
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  • How do you prove, "The area of the triangle is twice the product $xy$ where $y=f(x)$? – Kartal Tabak Feb 06 '21 at 21:02
  • I asked a separate question: https://math.stackexchange.com/questions/4015520/the-area-between-coordinate-axes-and-tangent-to-y-fx-is-2xy-is-this-cor. – Kartal Tabak Feb 06 '21 at 21:34
  • This answer has a correct calculation but a wrong explanation. You don't want to minimize $xy.$ You want to maximize $xy.$ It does happen to be true that once you find the maximum value of $xy,$ the area under the triangle tangent to the curve at $(x,y)$ will be $2xy.$ But at every other point on the curve the area of the triangle will be more than twice the product of the coordinates at the tangent point. – David K Feb 07 '21 at 02:55
  • @KartalTabak, have something here: https://math.stackexchange.com/questions/4015695/a-simple-method-to-calculate-minimal-area-enclosed-between-a-tangent-to-fx-a/4047395#4047395 – TurlocTheRed Mar 03 '21 at 17:17