Answer: The walking path around it.
Step-by-step explanation:
Remember the Pythagorean's theorem, for a triangle rectangle with catheti A and B, and with hypotenuse H, we have:
A^2 + B^2 = H^2
or
H = √(A^2 + B^2)
Here, A + B is the total distance for the path, while the hypotenuse is the distance for the bridge
Here we can see that:
A = 1170 ft
B = 520 ft
Then:
H = √( (520 ft)^2 + (1170 ft)^2) = 1280.4 ft
Now, let's compute the costs.
For the bridge, we know that each foot costs $11, then for 1280.4 ft the cost is:
Cost of the bridge = (1280.4)*$11 = $14,084.4
And for the walking path, the cost is $6 per foot, then the total cost of the path is:
Cost of the path = (520 ft + 1170 ft)*$6 = $10,140
We know that the bridge is preferred if it is within the range of $1500 for the path's cost.
This range is:
($10,140 - $1,500, $10,140 + $1,500) = ($8,640, $11,640)
Here we can see that the cost of the bridge does not belong to this range, (is higher) so the option we should recommend is the walking path around.
and differentiating both sides with respect to 
yields![\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B3x%5E2%2By%5E2%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B7%5Cbigg%5D%5Cimplies%206x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D0)
![\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B6x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B0%5Cbigg%5D%5Cimplies%206%2B2%5Cleft%28%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cright%29%5E2%2B2y%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D0)
and 
, we have