Math 220 March 7 I. If 1200 cm2 of material is available to make a ...
Math 220 March 7 I. If 1200 cm2 of material is available to make a ...
Math 220 March 7 I. If 1200 cm2 of material is available to make a ...
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m ′′ (x) = 360 − 60x 2<br />
m = 180x 2 − 5x 4<br />
m ′ (x) = 360x − 20x 3<br />
m ′ (x) = 20x(18 − x 2 )<br />
m ′′ (0) > 0<br />
m ′′ ( √ 18) < 0<br />
m ′′ ( √ 18) < 0<br />
0 = 20x(18 − x 2 )<br />
x = 0, ± √ 18<br />
There <strong>is</strong> a maximum at x = ± √ 18<br />
So the tangent line will have the largest slope at the points<br />
( √ 18, 1+ √ 18+60(18) 3/2 −(18) 5/2 ) and (− √ 18, 1+ √ 18+60(18) 3/2 −(18) 5/2 )<br />
IV. A farmer whats <strong>to</strong> fence in area <strong>of</strong> 400 square feet in a rectangular<br />
field and then divide it 3 sections with a two fence parallel <strong>to</strong> one <strong>of</strong> the sides<br />
<strong>of</strong> the rectangle. How can he do th<strong>is</strong> so as <strong>to</strong> minimize the cost <strong>of</strong> the fence?<br />
Answer:<br />
400 = xy<br />
Cost = 4x + 2y<br />
5