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Math Help - Maximum and minimum problems.

  1. #1
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    Maximum and minimum problems.

    1. A man in a row boat at point P, 5 km from the nearest point A on a straight shore wishes to reasvch point B, 6 km along the shore from point A, in the shortest time. Where should he land if he can row 2km/hr and walk 4km/hr (Hint: pick a point on the shore line)?

    2. An open rectangular tank (no top) with square ends is to have a volume of 6400 cubic meters. The base costs $75 per square meter and the sides $25 per square meter. What are the dimensions to build the tank at minimum cost?
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  2. #2
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    Max and min

    Hello iz1hp
    Quote Originally Posted by iz1hp View Post
    1. A man in a row boat at point P, 5 km from the nearest point A on a straight shore wishes to reasvch point B, 6 km along the shore from point A, in the shortest time. Where should he land if he can row 2km/hr and walk 4km/hr (Hint: pick a point on the shore line)?
    Suppose he lands at a point T, where AT = x km. Then, by Pythagoras, PT^2 = 25 + x^2

    \Rightarrow PT = \sqrt{25+x^2} km

    So this part of the journey takes ...?... hours.

    BT = ...?... km

    So this part takes ...?... hours.

    Add the two times together, to find the total time t. Then find the value of x that makes t a minimum.

    (Answer: x = \frac{5}{\sqrt{3}}= 2.89 km from A)
    2. An open rectangular tank (no top) with square ends is to have a volume of 6400 cubic meters. The base costs $75 per square meter and the sides $25 per square meter. What are the dimensions to build the tank at minimum cost?
    The ends of the tank are squares. Let's suppose they measure x meters by x meters. And suppose that the tank is y meters long. Then we have:

    Area of base = xy\, m^2, at $75 per m^2. So the cost of the base = $...?...

    The total area of all four sides is (2xy + 2x^2)\, m^2 at $25 per m^2. So the cost of the sides = $...?...

    So the total cost $C = ...?... (in terms of x and y)

    Now the volume of the tank = area of base x height = xy \times x = x^2y \,m^3.

    But this is 6400 \,m^3. So x^2y = 6400. So y = ...?... (in terms of x).

    Now express C in terms of x only, and find the value of x that will make C a minimum.

    (Answer: the tank measures 20 x 20 x 16 m.)

    Can you complete these now?

    Grandad
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