Based on the following information, what is the maximum width of the walkway?You are building a walkway of uniform width around a 100 ft by 60 ft swimming pool. After completing the walkway, you...

Based on the following information, what is the maximum width of the walkway?

You are building a walkway of uniform width around a 100 ft by 60 ft swimming pool. After completing the walkway, you want to put a fence along the outer edge of the walkway. You have 450 ft of fencing to enclose the walkway.

Asked on by catlover

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

The swimming pool is 100 by 60

Let a be the width of the walkway.

Then the perimeter is:

P1 = 2*100 + 2*60 = 320 ft.

However, the fence measures 450 ft.

Then we have a rectangle around the swimming pool whose perimeter is 450 ft.

Then:

Let x be the width and y be the length.

 p2 = 2*x + 2*y = 450.............(1)

==> x = 60 + 2a      

==> y= 100 + 2a   

Now substiute in (1)

==> 2(60+2a) + 2(100+2a) = 450

==> 120 + 4a + 200 + 4a = 450

==> 8a = 450 - 320

==> 8a = 130

==> a= 130/8

==> a= 16.25 ft

Then the width of the walkway is 16.25 ft.

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

Let:

Maximum width of the walkway = x ft.

Then the dimension of outer edge of walkway where the fencing is to be put can be calculated as:

Length of pool + walkway = Length of pool + 2(Width of walkway) = 100 + 2x ft.

Width of pool + walkway = Width of pool + 2(Width of walkway) = 60 + 2x ft.

Then:

Total perimeter of outside edge of the walkway = 2(Length + Width)

= 2(100 + 2x + 60 + 2x)

= 320 + 8x

Maximum value of this is given as 450. There maximum value of x will be given when the above expression for length is equated to 450. Doing this we get:

8x + 320 = 450

==> 8x = 450 - 320

==> 8x = 130

==> x = 130/8 = 16.25

Answer:

Maximum width of walkway = 16.25 ft.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Here  the  width h of way  around the swimming pool is uniform width  is a variable.

So we give you the solution for the maximum h in such a way that the material of 450 ft fencing is fully used.

Let h be the width of the path. Then the length and width of the fence = 100+2h and 60+2h.

So the perimeter of the fencing  of rectangle  around the swimming pool = 2{100+2h+60+2h} = 320+8h ft which should be the length  450 ft of the fencing material .

Therefore,  320+8h = 450

8h = 450-320 = 130

Therefore h = 16.25 ft.

Therefore the length of the fence = 100+2h = 100+2*16.25 = 132.5 .

The width of the fence = 60+2h = 60+2*16.25 = 92.5.

So the maximum width  of the way around =  16.25 feet.

 

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