a rectangular flower garden with an area of 30 m² is surrounded by a grass border 1 m wide on two sides and…

Explanation:

Step1: Let the dimensions of the flower - garden

Let the length of the flower - garden be $x$ meters and the width be $y$ meters. We know that the area of the flower - garden is $A = xy=30$, so $y=\frac{30}{x}$.

Step2: Find the combined area function

The combined area $A_{total}$ of the garden and the borders: The length of the combined region is $(x + 4)$ meters and the width is $(y + 2)$ meters. So $A_{total}=(x + 4)(y + 2)$. Substitute $y=\frac{30}{x}$ into the equation: $A_{total}=(x + 4)(\frac{30}{x}+2)$.

Step3: Expand the function

Expand $(x + 4)(\frac{30}{x}+2)$: [ \begin{align*} A_{total}&=x\cdot\frac{30}{x}+2x + 4\cdot\frac{30}{x}+4\times2\ &=30+2x+\frac{120}{x}+8\ &=38 + 2x+\frac{120}{x} \end{align*} ]

Step4: Find the derivative

Differentiate $A_{total}(x)$ with respect to $x$. Using the power rule, if $f(x)=ax^n$, then $f^\prime(x)=nax^{n - 1}$. So $A_{total}^\prime(x)=2-\frac{120}{x^{2}}$.

Step5: Set the derivative equal to zero

Set $A_{total}^\prime(x)=0$: [ \begin{align*} 2-\frac{120}{x^{2}}&=0\ 2&=\frac{120}{x^{2}}\ 2x^{2}&=120\ x^{2}&=60\ x&=\sqrt{60}=2\sqrt{15} \end{align*} ]

Step6: Find the second - derivative

Differentiate $A_{total}^\prime(x)$ to get the second - derivative $A_{total}^{\prime\prime}(x)=\frac{240}{x^{3}}$. When $x = 2\sqrt{15}$, $A_{total}^{\prime\prime}(x)>0$, so the function has a minimum at this point.

Step7: Find the value of $y$

Since $y=\frac{30}{x}$, when $x = 2\sqrt{15}$, $y=\frac{30}{2\sqrt{15}}=\sqrt{15}$.

Answer:

$\sqrt{15}$