Question

A rectangular piece of land is to be fenced using two kinds of fencing. Two opposite sides will be fenced using standard fencing that costs $6 m, while the other two sides will require heavy-duty fencing that costs $9 m. What are the dimensions of the rectangular lot of greatest area that can be fenced for a cost of $9000?

Answer 1

Answer:

Therefore the dimension of the rectangular piece of land is 375 m by 250 m.

Step-by-step explanation:

Given that,

Two opposite sides of rectangular land will be fenced using standard fencing that costs $6 per m while the other two side will require heavy duty fencing that costs $9 per m.

Let the length and width of the rectangular x and y respectively.

Then total cost for fencing = $\ (2x\times 6+2y\times 9)$

                                            =$(12x+18y)

Then,

12x+18y= 9,000

⇒ 2x+3y=1,500     [ cancel out common factor 6]

⇒3y= 1500 - 2x

$\Rightarrow y=\frac{1500-2x}{3}$ ....(1)

The area of the rectangular land is = Length × width

                                                          = xy

Let,

A=xy

Putting the value of y

$A=x.\frac{1500-2x}{3}$

$\Rightarrow A=\frac13(1500x-2x^2)$

Differentiating with respect to x

$A'=\frac13(1500-4x)$

Again differentiating with respect to x

$A''=-\frac43$

Now we have to determine the critical value of A.

For critical value set A'=0

$\frac13(1500-4x)=0$

⇒1500-4x=0

$\Rightarrow x=\frac{1500}{4}$

$\Rightarrow x=375$ m

Since, $A''|_{x=375}=-\frac43$. at x=375 m the area of the rectangular piece of land is maximum.

From equation (1) we get when x=375

$y=\frac{1500-2\times 375}{3}$

⇒y=250 m

Therefore the dimension of the rectangular piece of land is 375 m by 250 m.