A rectangular recreational field needs to be built outside of a gymnasium. Three walls of fencing are needed and the fourth wall is to be a wall of the gymnasium itself. The ideal area for such a field is exactly 90000ft2. In order to minimize costs, it is necessary to construct the fencing using the least amount of material possible. Assuming that the material used in the fencing costs $49/ft, what is the least amount of money needed to build this fence of ideal area? Round your answer to the nearest two decimal places.

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DevonFen DevonFen · Answer 1 · 2020-04-10T09:59:09+02:00

Answer:

$41,577.88 is needed to build this fence of ideal area.

Step-by-step explanation:

Given that, the area of the rectangular field is 90000 ft².

Let the of the length of the field which side is also a wall of the gymnasium be x and the width of the field be y.

Therefore,

xy=90000

[tex]\Rightarrow y=\frac{90000}{x}[/tex]

The length of fencing of the field = 2(x+y)-x

= x+2y

The material cost for the fencing is $ 49 per ft.

Total cost for the fencing is = 49(x+2y)

Let,

C=49(x+2y)

Putting [tex]y=\frac{90000}{x}[/tex] in the above equation

[tex]C=49(x+2\times\frac{90000}{x})[/tex]

[tex]\Rightarrow C=49(x+\frac{180000}{x})[/tex] ........(1)

Differentiating with respect to x

[tex]C'=49(1-\frac{180000}{x^2})[/tex]

Again differentiating with respect to x

[tex]C''=49(\frac{360000}{x^3})[/tex]

Now set C'=0

[tex]49(1-\frac{180000}{x^2})=0[/tex]

[tex]\Rightarrow \frac{180000}{x^2}=1[/tex]

[tex]\Rightarrow x^2=180000[/tex]

[tex]\Rightarrow x=424.26[/tex] ft.

[tex]C''|_{x=424.26}=49\times \frac{360000}{(424.26)^3}>0[/tex]

Since at x= 424.26, C''>0. So at x=424.26,the value of C will be maximum.

Putting x= 424.26 in equation (1)

[tex]C=49(424.26+\frac{180000}{424.26})[/tex]

=$41,577.88

$41,577.88 is needed to build this fence of ideal area.