Chapter 4.6, Problem 50E

To determine

To calculate:The frame of a kite is to be made from six pieces of wood. The four exterior pieces have been cut with lengths indicated in the figure. To maximize the area of the kite, how long should the diagonal pieces be?

Answer to Problem 50E

The shortest diagonal piece is $\frac{2ab}{\sqrt{a^2+b^2}}$ and the longer diagonal piece is $\sqrt{a^2+b^2}$

Explanation of Solution

Given information:

The frame of a kite is to be made from six pieces of wood. The four exterior pieces have been cut with lengths indicated in the figure

Formula used:

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

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

Product rule: If the two function $f\left(x\right)\text{ and }g\left(x\right)$ are differentiable then the product is differentiable and ,

  ${\left(fg\right)}^{\prime }=f{g}^{\prime }+g{f}^{\prime }$

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. $f^{\prime }(x)$ changessign 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

Draw the diagram frame of a kite is to be made from six pieces of wood. The four exterior pieces have been cut with lengths indicated in the figure,

  

If $x$ be the half the diagonal of kite, Then

Recall that,

From the figure,

Area of right angled triangle:

  $\begin{array}{l}A\left(x\right)=\frac{1}{2}\times \text{base}\times \text{perpendicular}\\ =2\left(\frac{1}{2}\times x\times \sqrt{a^2-x^2}\right)+2\left(\frac{1}{2}\times \sqrt{a^2-x^2}\times \sqrt{x^2+b^2-a^2}\right)\\ =\sqrt{a^2-x^2}\left(x+\sqrt{x^2+b^2-a^2}\right)\end{array}$ y

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 $f^{\prime }(a)=0$ and
1. $f^{\prime }(x)$ 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 $x$

  $A^{\prime }\left(x\right)=\frac{d}{dx}\left\{\sqrt{a^2-x^2}\left(x+\sqrt{x^2+b^2-a^2}\right)\right\}$

Recall that,

Product rule: If the two function $f\left(x\right)\text{ and }g\left(x\right)$ are differentiable then the product is differentiable and ,

  ${\left(fg\right)}^{\prime }=f{g}^{\prime }+g{f}^{\prime }$

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

  $\begin{array}{l}\sqrt{a^2-x^2}\left(1+\frac{x}{\sqrt{x^2+b^2-a^2}}\right)-\frac{x}{\sqrt{a^2-x^2}}\left(x+\sqrt{x^2+b^2-a^2}\right)=0\\ \Rightarrow \frac{x}{\sqrt{a^2-x^2}}=\frac{\sqrt{a^2-x^2}}{\sqrt{x^2+b^2-a^2}}\\ \Rightarrow x=\frac{a^2}{\sqrt{a^2+b^2}}\end{array}$

Evaluate the $A\left(x\right)$ at the critical point and end point

At $x=0$

  $\begin{array}{l}A\left(0\right)=\sqrt{a^2-0^2}\left(0+\sqrt{0^2+b^2-a^2}\right)\\ =a\sqrt{b^2-a^2}\end{array}$

At $x=a$

  $\begin{array}{l}A\left(a\right)=\sqrt{a^2-a^2}\left(a+\sqrt{a^2+b^2-a^2}\right)\\ =0\end{array}$

At $x=\frac{a^2}{\sqrt{a^2+b^2}}$

  $\begin{array}{l}A\left(a\right)=\sqrt{a^2-{\left(\frac{a^2}{\sqrt{a^2+b^2}}\right)}^2}\left(\frac{a^2}{\sqrt{a^2+b^2}}+\sqrt{{\left(\frac{a^2}{\sqrt{a^2+b^2}}\right)}^2+b^2-a^2}\right)\\ A\left(a\right)=\sqrt{a^2-\frac{a^4}{a^2+b^2}}\left(\frac{a^2}{\sqrt{a^2+b^2}}+\sqrt{\frac{a^4}{a^2+b^2}+b^2-a^2}\right)\\ =ab\end{array}$

Therefore, there is an absolute maximum at $x=\frac{a^2}{\sqrt{a^2+b^2}}$

So, the shortest diagonal piece is

  $=\frac{2ab}{\sqrt{a^2+b^2}}$

The longer diagonal piece is

  $=\sqrt{a^2+b^2}$

Conclusion:

Thus the shortest diagonal piece is $\frac{2ab}{\sqrt{a^2+b^2}}$ and the longer diagonal piece is $\sqrt{a^2+b^2}$