Question

A rectangular garden of area 480 square feet is to be surrounded on three sides by a brick wall costing $ 12 per foot and on one side by a fence costing $ 8 per foot. Find the dimensions of the garden such that the cost of the materials is minimized

Answer 1

Answer:

Therefore the dimensions of the rectangular garden are 23.42 ft by 20.5 ft.

Step-by-step explanation:

Given that,

A rectangular garden of area 480 square feet.

Let length of the rectangular garden be x which is surrounded by fence and width of the rectangular garden be y.

Then xy is the area of the given rectangular garden .

Then,

xy= 480

$\Rightarrow y=\frac{480}{x}$

The length of the tree sides which are surrounded by brick wall is = 2y+x.

The cost for the brick wall is =Length×cost per feet= $12(2y+x)

The cost for the fencing is =Length×cost per feet= $ 9x

$\therefore C=12(2y+x)+9x$

Now putting $y=\frac{480}{x}$

$\therefore C=12(2.\frac{480}{x}+x)+9x$

$\Rightarrow C=\frac{11520}{x}+21x$

Differentiating with respect to x

$C'=-\frac{11520}{x^2}+21$

Again differentiating with respect to x

$C''=\frac{23040}{x^3}$

Now we set C'=0

$\therefore-\frac{11520}{x^2}+21=0$

$\Rightarrow\frac{11520}{x^2}=21$

$\Rightarrow x^2=\frac{11520}{21}$

$\Rightarrow x\approx 23.42$

$C''|_{x=23.42}=\frac{23040}{(23.42)^3}>0$.

Since at x=23.42,C''>0. So at x=23.42, the total cost will be minimum.

The width of the rectangular garden is $y=\frac{480}{x}$

                                                                    $=\frac{480}{23.42}$

                                                                    $\approx 20.5$

Therefore the dimensions of the rectangular garden are 23.42 ft by 20.5 ft.

The cost of the material is $C=\frac{11520}{x}+21x$

                                                 $=\frac{11520}{23.42}+21\times 23.42$

                                                 =$983.70