Problem 1.Initially, let ship B be at the origin. Then ship A is initially at the point (100,0). So at any time t, ship B is at the point (27t, 0) and ship A is at the point (100, 34t).So the distance between the two ships may be determined by the Pythagorean Theorem, which leads to the equation below for d, the distance between the two ships:d := sqrt((34*t)^2 + (100-27*t)^2);

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The distance between the ships is not changing when the derivative of d is 0:dd := diff(d,t);

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solve(dd=0,t);

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hrs := evalf(%);

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mins := 60*(hrs-1);

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So at 2:26 pm (i.e., one hour and 26 minutes after 1 pm), the distance between the ships is not changing.plot(d, t=0..3);

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Problem 2.f := (cos(3*x+Pi/3))^4;

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q := int(f,x);

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We need a "plus c" to get the general antiderivative:F := q + c;

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We want x = Pi/6:subs(x=Pi/6,F);

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simplify(%);

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What is c when y = 2?cc := solve(%=2, c);

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finalF := subs(c=cc, F);

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In this last line is y as a function of x.Problem 3.h := 8 + 5/12;

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w := 23 + 10/12;

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Draw the following picture: A right triangle, the base of which is the ground (length x+w), the height of which is part of the building (length y), and whose hypotenuse is B, the length of the beam. Now draw in the wall, parallel to the building, with height h.Via similar triangles:y := h * (x+w) / x;

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Via Pythagorean Theorem:B := simplify(sqrt((x+w)^2 + y^2));

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q := 'q':plot(B,x=0..30,q=0..100);

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B is maximized when its derivative is 0:dB := simplify(diff(B,x));

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solve(dB=0,x);

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evalf(%);

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Pick the real solution:%[1];

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What's the value of B at this solution?subs(x=%,B);

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So 43 feet and...%-43;

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%*12;

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... 9 inches is the maximum length for the beam.