### Problem StatementTwo poles are connected by a wire that is also connected to the ground. The first pole is 20 ft. tall and the second pole is 10 ft. tall. There is a 30 ft. distance between the two poles. Where should the wire be anchored to the ground between the poles to minimize the amount of wire needed?### Diagram Explanation- There is a horizontal line representing the ground.- The first pole is on the left, labeled with a height of 10 ft.- The second pole is on the right, labeled with a height of 20 ft.- The distance between the two poles along the ground is marked as 30 ft.- A point on the ground is marked as "x" from the first pole, indicating where the wire might be anchored.- Two angled lines represent the wire segments connecting the tops of the poles to the point on the ground.### ObjectiveDetermine the optimal position "x" where the wire should be anchored to minimize its total length.

Given: Two poles are connected by a wire that is also connected to the ground.The first pole is 20 ft. tall and the second pole is 10 ft tall.There is 30 ft. distance between the two poles. To Find: Where should the wire be anchored to the ground between the polesto minimize the amount of wire needed.Explanation: Let the wire anchored to the ground at x ft. from the left pole. from the figure we have, the amount of wire is:L=a+b.............iNow, we will calculate a and b using Pythagoras theoremb2=202+30-x2⇒b2=400+900+x2-60x⇒b2=x2-60x+1300...........iia2=102+x2=100+x2a2=x2+100...........iiiNow, from i, ii, and iii, we haveL=x2+100+x2-60x+1300.......ivNow, we will use first derivative test to find minimum amount of wire neededDifferentiating both side of iv, with respect to x, we getL'(x)=2x2x2+100+2x-602x2-60x+1300⇒L'(x)=xx2+100+x-30x2-60x+1300⇒L'(x)=xx2-60x+1300+x-30x2+100x2-60x+1300x2+100now, put L'(x)=0, we getxx2-60x+1300+x-30x2+100=0⇒xx2-60x+1300=30-xx2+100squaring both side, we get:xx2-60x+13002=30-xx2+1002⇒x2x2-60x+1300=30-x2x2+100⇒x4-60x3+1300x2=900+x2-60xx2+100⇒x4-60x3+1300x2=900x2+x4-60x3+90000+100x2-6000x⇒300x2+6000x-90000=0⇒x2+20x-300=0⇒x2+30x-10x-300=0⇒xx+30-10x+30=0⇒x=10x≠-30, ∵distance can't be negativehence, the wire anchored at a distance of x=10 ft. from the left pole.Answer: The wire anchored at a distance of x=10 ft. from the left pole.