# Solving Optimization Problems

## Key Questions

• Let $x$ and $y$ be the base and the height of the rectangle, respectively.

Since the area is 100 ${m}^{2}$,

$x y = 100 R i g h t a r r o w y = \frac{100}{x}$

The perimeter $P$ can be expressed as

$P = 2 \left(x + y\right) = 2 \left(x + \frac{100}{x}\right)$

So, we want to minimize $P \left(x\right)$ on $\left(0 , \infty\right)$.

By taking the derivative,

$P ' \left(x\right) = 2 \left(1 - \frac{100}{x} ^ 2\right) = 0 R i g h t a r r o w x = \pm 10$

$x = 10$ is the only critical value on $\left(0 , \infty\right)$

$y = \frac{100}{10} = 10$

By testing some sample values,

$P ' \left(1\right) < 0 R i g h t a r r o w P \left(x\right)$ is decreasing on $\left(0 , 10\right]$.

$P ' \left(11\right) > 0 R i g h t a r r o w P \left(x\right)$ is increasing on $\left[10 , \infty\right)$

Therefore, $P \left(10\right)$ is the minimum

I hope that this was helpful.

Hence, the dimensions are $10 \setminus \times 10$.

The dimensions of the rectangle is $\sqrt{2} r$ and $\frac{r}{\sqrt{2}}$

#### Explanation:

The equation of the semicircle is

${x}^{2} + {y}^{2} = {r}^{2}$.......................$\left(1\right)$

The area of the rectangle is

$A = 2 x y$....................$\left(2\right)$

From equation $\left(1\right)$, we get

${y}^{2} = {r}^{2} - {x}^{2}$

$y = \sqrt{{r}^{2} - {x}^{2}}$

Plugging this value in equation $\left(2\right)$

$A = 2 x \sqrt{{r}^{2} - {x}^{2}}$

Differentiating wrt $x$ using the product rule

$\frac{\mathrm{dA}}{\mathrm{dx}} = 2 \sqrt{{r}^{2} - {x}^{2}} - 2 {x}^{2} / \sqrt{{r}^{2} - {x}^{2}}$

$= \frac{2 {r}^{2} - 2 {x}^{2} - 2 {x}^{2}}{\sqrt{{r}^{2} - {x}^{2}}}$

$= \frac{2 {r}^{2} - 4 {x}^{2}}{\sqrt{{r}^{2} - {x}^{2}}}$

The critical points are when

$\frac{\mathrm{dA}}{\mathrm{dx}} = 0$

That is

$\frac{2 {r}^{2} - 4 {x}^{2}}{\sqrt{{r}^{2} - {x}^{2}}} = 0$

${r}^{2} = 2 {x}^{2}$

$x = \frac{r}{\sqrt{2}}$

Then,

$y = \sqrt{{r}^{2} - {x}^{2}} = \sqrt{{r}^{2} - {r}^{2} / 2} = \frac{r}{\sqrt{2}}$

The maximum area is

$A = 2 \cdot \frac{r}{\sqrt{2}} \cdot \frac{r}{\sqrt{2}} = {r}^{2}$

• Let $\left(x , y\right)$ be a point on the ellipse $4 {x}^{2} + {y}^{2} = 4$.

$\Leftrightarrow {y}^{2} = 4 - 4 {x}^{2} \Leftrightarrow y = \pm 2 \sqrt{1 - {x}^{2}}$

The distance $d \left(x\right)$ between $\left(x , y\right)$ and $\left(1 , 0\right)$ can be expressed as

$d \left(x\right) = \sqrt{{\left(x - 1\right)}^{2} + {y}^{2}}$

by ${y}^{2} = 4 - 4 {x}^{2}$,

$= \sqrt{{\left(x - 1\right)}^{2} + 4 - 4 {x}^{2}}$

by multiplying out

$= \sqrt{- 3 {x}^{2} - 2 x + 5}$

Let us maximize $f \left(x\right) = - 3 {x}^{2} - 2 x + 5$

$f ' \left(x\right) = - 6 x - 2 = 0 R i g h t a r r o w x = - \frac{1}{3}$ (the only critical value)

$f ' ' \left(x\right) = - 6 R i g h t a r r o w x = - \frac{1}{3}$ maximizes $f \left(x\right)$ and $d \left(x\right)$

Since $y = \pm 2 \sqrt{1 - {\left(- \frac{1}{3}\right)}^{2}} = \pm \frac{4 \sqrt{2}}{3}$,

the farthest points are $\left(- \frac{1}{3} , \pm \frac{4 \sqrt{2}}{3}\right)$.