4.4 Optimization

Download Report

Transcript 4.4 Optimization

4.4 Optimization

Strategies for solving Max and Min Story Problems: 1.

2.

3.

4.

5.

Example 1: What is the maximum product of two non negative numbers whose sum is 20?

Example 2 A rancher wishes to fence off a rectangular pasture along a straight river, the side along the river requiring no fence. She has enough barbed wire to build a fence 6000 feet long. What are the dimensions of the pasture which gives the largest grazing area?

Example 3: An open-top box is to be made by cutting congruent squares from the corners of a 20 by 25 inch sheet of tin and bending up the sides. How large should the squares be make the box hold as much as possible? What is the resulting maximum volume?

Example 4: Find two numbers whose sum is 20 and product is as large as possible .

What is the largest possible area for a right triangle whose hypotenuse is 5 cm long , and what are its dimensions?

Answers on Next Slide…

Answers 𝐴 = 1 2 π‘₯ 25 βˆ’ π‘₯ 2 𝐷: [0, 5] 𝐴 = βˆ’ 1 2 π‘₯ 2 25 βˆ’ π‘₯ 2 βˆ’ 1 2 + 1 2 25 βˆ’ π‘₯ 2 𝐢𝑃: π‘₯ = 3.5355

𝑆𝑖𝑛𝑐𝑒 π‘“π‘–π‘Ÿπ‘ π‘‘ π‘‘π‘’π‘Ÿπ‘–π‘£π‘Žπ‘‘π‘–π‘£π‘’ π‘β„Žπ‘Žπ‘›π‘”π‘’π‘  π‘“π‘Ÿπ‘œπ‘š π‘π‘œπ‘  π‘‘π‘œ 𝑛𝑒𝑔 π‘Žπ‘‘ x = 3.5355, by 1DT it’s a rel max. Since it’s the only CP, it’s an Abs Max. Dimension: 3.5355 x 3.5355

Area = 6.25

A 216 m 2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed?

Answers on Next Slide…

Answers 𝑃 = 3x βˆ’ 432π‘₯ βˆ’1 𝐷: [0, ∞) 𝑃′ = 3 βˆ’ 432π‘₯ βˆ’2 𝐢𝑃: π‘₯ = 12 P” = 864x -3 P”(12) > 0 so it’s a rel max Since it’s the only CP it is an absolute max

The US Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around), as shown in the figure, does not exceed 108 in. What dimensions will give a box with a square end the largest possible volume?

Girth = Distance around here square end length

Answers 𝑉 = 108π‘₯ 2 βˆ’ 4π‘₯ 3 𝐷: [0, 27] 𝑉 β€² = 216π‘₯ βˆ’ 12π‘₯ 2 𝐢𝑃: π‘₯ = 0, 18

x

0 18 27

V

0 11664 0 π·π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘ : π‘₯ = 18, 𝑦 = 36