# Thread: Maximum and minimum problems.

1. ## 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?

2. ## Max and min

Hello iz1hp
Originally Posted by iz1hp
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 $\displaystyle AT = x$ km. Then, by Pythagoras, $\displaystyle PT^2 = 25 + x^2$

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

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

$\displaystyle BT =$ ...?... km

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

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

(Answer: $\displaystyle x = \frac{5}{\sqrt{3}}= 2.89$ 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 $\displaystyle x$ meters by $\displaystyle x$ meters. And suppose that the tank is $\displaystyle y$ meters long. Then we have:

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

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

So the total cost $C = ...?... (in terms of$\displaystyle x$and$\displaystyle y$) Now the volume of the tank = area of base x height =$\displaystyle xy \times x = x^2y \,m^3$. But this is$\displaystyle 6400 \,m^3$. So$\displaystyle x^2y = 6400$. So$\displaystyle y =$...?... (in terms of$\displaystyle x$). Now express$\displaystyle C$in terms of$\displaystyle x$only, and find the value of$\displaystyle x$that will make$\displaystyle C\$ a minimum.

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

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# a person being in a boat 'a'km from the nearest point of the beach wishes to reach as quickly as possible a point b km from that point along the shore

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