Chapter 4.6, Problem 42E

To determine

To calculate: The smallest possible area of the triangle that is cut off by the first quadrant and whose hypotenuse is tangent to the parabola $y=4-x^2$ at some point.

Answer to Problem 42E

The smallest possible area is $A\approx 6.15$

Explanation of Solution

Given information:

The triangle that is cut off by the first quadrant and whose hypotenuse is tangent to the parabola $y=4-x^2$ at some point

Formula used:

Area of right angled triangle: $=\frac{1}{2}\times \text{base}\times \text{perpendicular}$

Point-Slope form of the equation of a line is

  $y-y_1=m\left(x-x_1\right)$

Where $m$ is the slope and $\left(x_1,y_1\right)$ is the point in the line.

Quadratic formula: let the quadratic equation is $a{x}^2+bx+c=0,\left(a\ne 0\right)$ Thus, if $b^2-4ac\ge 0$ then the roots of the quadratic equation is given by

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$
   , if$f^{\prime }(a)=0$ and
   1. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

• If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
• If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

The triangle that is cut off by the first quadrant and whose hypotenuse is tangent to the parabola $y=4-x^2$ at some point

Let $x=a$ be where the line segment is tangent to the parabola,

The parabola

  $y=4-x^2\mathrm{......}\left(1\right)$

Substitute $x=a$ to get,

  $y=4-a^2$

For slope, differentiate equation $\left(1\right)$ with respect to $x$

  $y^{\prime }=-2x$

Let us denote the slope of tangent as $M\left(x\right)$

  $M\left(x\right)=-2x$

An equation of the line that passing through $\left(a,4-a^2\right)$ with slope $M\left(x\right)=-2x$ is

Recall that,

Point-Slope form of the equation of a line is

  $y-y_1=m\left(x-x_1\right)$

Where $m$ is the slope and $\left(x_1,y_1\right)$ is the point on the line

  $\begin{array}{l}y-\left(4-a^2\right)=-2a\left(x-a\right)\\ \Rightarrow y-4+a^2=-2ax+2a^2\\ \Rightarrow y=a^2-2ax+4\mathrm{......}\left(2\right)\end{array}$

Substitute $x=0$ in equation $\left(2\right)$ to get $y-\text{intercept}$

  $y=a^2+4$

Substitute $y=0$ in equation $\left(2\right)$ to get $x-\text{intercept}$

  $\begin{array}{l}0=a^2-2ax+4\\ \Rightarrow 2ax=a^2+4\\ \Rightarrow x=\frac{a}{2}+\frac{2}{a}\\ \Rightarrow x=\frac{a^2+4}{2a}\end{array}$

Recall that,

Area of right angled triangle: $=\frac{1}{2}\times \text{base}\times \text{perpendicular}$

  $\begin{array}{l}A\left(a\right)=\frac{1}{2}\times \left(\frac{a^2+4}{2a}\right)\times \left(a^2+4\right)\\ =\frac{1}{4}\left(a^3+8a+\frac{16}{a}\right)\\ =\frac{1}{4}{a}^3+2a+\frac{4}{a}\end{array}$

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

• If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
• If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate with respect to $a$

  $A^{\prime }\left(a\right)=\frac{3}{4}{a}^2+2-\frac{4}{a^2}\mathrm{......}\left(3\right)$

Solve for $A^{\prime }\left(a\right)=0$ , and simplified

  $\begin{array}{l}\frac{3}{4}{a}^2+2-\frac{4}{a^2}=0\\ \Rightarrow 3a^4+8a-16=0\end{array}$

Recall that,

Quadratic formula: let the quadratic equation is $a{x}^2+bx+c=0,\left(a\ne 0\right)$ Thus, if $b^2-4ac\ge 0$ then the roots of the quadratic equation is given by

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

  $\begin{array}{l}3a^4+8a-16=0\\ a^2=\frac{-8\pm \sqrt{8^2-4\times 3\times -16}}{2\times 3}\\ =\frac{-8\pm \sqrt{256}}{6}\\ =\frac{-8\pm 16}{6}\end{array}$

Solve for different sign’s

  $\begin{array}{l}{a}^2=\frac{-8+16}{6}\text{or}{a}^2=\frac{-8-16}{6}\\ a^2=\frac{8}{6}\text{or}{a}^2=-\frac{24}{6}\\ a^2=\frac{4}{3}\text{or}{a}^2=-4\\ a=\frac{2}{\sqrt{3}}\text{or}a=\sqrt{-4}\end{array}$

Take positive root for $a$ because to stay in the first quadrant

Differentiate equation $\left(3\right)$ with respect to $a$

  $A^{″}\left(a\right)=\frac{6}{4}a+\frac{8}{a^3}$

Therefore,

  $A^{″}\left(a\right)=\frac{6}{4}a+\frac{8}{a^3}\ge 0$ at $a=\frac{2}{\sqrt{3}}$ , hence there is local minimum at $a=\frac{2}{\sqrt{3}}$

Now put $a=\sqrt{3}$ in $A\left(a\right)=\frac{1}{4}{a}^3+2a+\frac{4}{a}$ and solve

  $\begin{array}{l}A\left(\frac{2}{\sqrt{3}}\right)=\frac{1}{4}{\left(\frac{2}{\sqrt{3}}\right)}^3+2\left(\frac{2}{\sqrt{3}}\right)+\frac{4}{\left(\frac{2}{\sqrt{3}}\right)}\\ =\frac{2\sqrt{3}+12\sqrt{3}+18\sqrt{3}}{9}\\ =\frac{32\sqrt{3}}{9}\\ \approx 6.15\end{array}$

Conclusion:

Thus the smallest possible area is $A\approx 6.15$