# optimization problem minimizing cost

• Sep 10th 2010, 05:56 AM
euclid2
optimization problem minimizing cost
A new cottage is built across the river and 300m downstream from the nearest telephone relay station. the river is 120m wide. in order to wire the cottage for phone service, wire will be laid across the river under water, and along the edge of the river above ground. the cost to lay wire under water is 15$/m and the cost to lay above ground is 10$/m. how much wire should be laid underwater to minimize the cost?

i set up total cost=

T(c)={\frac{\sqrt{{x^2+120^2}}{15}} + {\frac{300-x}{10}}

T(c)=\frac{1}{15}(x^2+14400)^{\frac{1}{2}} + \frac{1}{10}(300-x)

then i differentiated to get

T'(c)= \frac{1}{15} \frac{1}{2} (x^2+14400)^{{\frac{-1}{2}}(2x) - \frac{1}{10}

T'(c)=\frac{x}{(15(x^2+14400)^{\frac{1}{2} - \frac{1}{10}

i set T'(c)=0 and solved and got a negative number which i could not take the square root of which is where i know i went wrong. where did i go wrong? having trouble with LaTex

i think the problem may be that the original equation made up should be

T(c)={\frac{\sqrt{{x^2+120^2}}{15}} (-) {\frac{300-x}{10}}

• Sep 10th 2010, 07:03 AM
euclid2
forget the work in the first post i have re-looked the problem and came up with this:

$T(c)= \sqrt{x^2+120^2}(15)+(300-x)(10)$

i expanded and differentiated to get

$T'(c)=\frac{15x}{(x^2+14400)^\frac{1}{2}} - 10$

so i need to solve

$\frac{15x}{(x^2+14400)^\frac{1}{2}} - 10 = 0$

how do i do this?
• Sep 10th 2010, 10:45 AM
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Quote:

Originally Posted by euclid2
A new cottage is built across the river and 300m downstream from the nearest telephone relay station. the river is 120m wide. in order to wire the cottage for phone service, wire will be laid across the river under water, and along the edge of the river above ground. the cost to lay wire under water is 15$/m and the cost to lay above ground is 10$/m. how much wire should be laid underwater to minimize the cost?

For this type of problem I like to draw a diagram. Excuse the mouse + MS paint ugliness.

http://www.mathhelpforum.com/math-he...8&d=1284144287

So we want to find the value of H that minimizes cost.

D^2 + 120^2 = H^2
C = 15H + 10(300-D)

Use the first equation to substitute into the second so that you have C in terms of H, then differentiate and set to 0. (For completeness, I should mention that in general also don't forget to check the endpoints, here given by D = 0 and D = 300.)

$C(H)=15H+10(300-\sqrt{H^2-120^2})$

I see you solved for C in terms of D instead of H (taking into account the different names of variables); this will add an extra step on at the end.

Since I wrote it out with my variable names, it's

$\frac{dC(D)}{dD} = \frac{15D}{\sqrt{D^2+14400}}-10=0$

Continuing,

$10=\frac{15D}{\sqrt{D^2+14400}}$

$10\sqrt{D^2+14400}=15D$

$\sqrt{D^2+14400}=\frac{3D}{2}$

$D^2+14400=\frac{9D^2}{4}$

$5D^2=57600$

$D=\sqrt{\frac{57600}{5}}=\sqrt{11520}=48\sqrt{5}$

And then of course don't forget to solve for H.