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

a rectangular flower garden with an area of 30 m² is surrounded by a grass border 1 m wide on two sides and a 2 m wide on the other two sides as shown in the figure. what dimensions of the garden minimize the combined area of the garden and borders? the shorter edge (rectangular height) of the garden that will minimize the total area is √15 m. (type an exact answer, using radicals as needed.) the longer edge (rectangular width) of the garden that will minimize the total area is m. (type an exact answer, using radicals as needed.)

a rectangular flower garden with an area of 30 m² is surrounded by a grass border 1 m wide on two sides and a 2 m wide on the other two sides as shown in the figure. what dimensions of the garden minimize the combined area of the garden and borders? the shorter edge (rectangular height) of the garden that will minimize the total area is √15 m. (type an exact answer, using radicals as needed.) the longer edge (rectangular width) of the garden that will minimize the total area is m. (type an exact answer, using radicals as needed.)

Answer

Explanation:

Step1: Let the shorter - side of the garden be $x$ and the longer - side be $y$.

Given the area of the flower garden is $A = xy=30$, so $y=\frac{30}{x}$.

Step2: Calculate the combined area $A_{total}$.

The combined area of the garden and the borders: $A_{total}=(x + 2)(y+4)$. Substitute $y = \frac{30}{x}$ into it, we get $A_{total}=(x + 2)(\frac{30}{x}+4)=30+4x+\frac{60}{x}+8=38 + 4x+\frac{60}{x}$.

Step3: Find the derivative of $A_{total}$ with respect to $x$.

$A_{total}'(x)=4-\frac{60}{x^{2}}$.

Step4: Set the derivative equal to zero to find critical points.

$4-\frac{60}{x^{2}} = 0$. Then $4x^{2}-60 = 0$, so $x^{2}=15$, and $x=\sqrt{15}$ (we take the positive value since $x$ represents a length).

Step5: Find the value of $y$.

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

Answer:

$2\sqrt{15}$