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Unformatted text preview: 8. cardboard ft maximize volume base side x
cut out squares feet on a side. by by x dimensions of box are so volume is Since all dimensions must be positive, we know that . Thus we are to maximize in this interval. At the end points, we have we determine critical numbers in the interval from the derivative: critical numbers or at we have ft MTH 173 section 4.7 notes, page 1 the maximum volume of 2 ft occurs when squares of each corner are cut out from MTH 173 section 4.7 notes, page 2 10. box with square base volume 32,000 cm . minimize material To minimize the material use requires we minimize the surface area of the box, that is, we minimize he total area of the base and the 4 sides of the box. let the length of a side of the square base, height. Since volume surface area determine to minimize determine such that so is critical number when at is minimized at so dimensions are 40 by 40 by 20 cm MTH 173 section 4.7 notes, page 3 16. closest point on to from to where this is a request to minimize the distance , but replace with to give minimize when is only critical number since , at there is a minumum value of everywhere, at the critical number is the point on the line closest to MTH 173 section 4.7 notes, page 4 22. dimensions of rectangle of max area with base on vertices on axis and other 8 6 (x,y)
4 here the base of the yellow rectangle has length , while the height is , so the area is but , so 2 -3 -2 -1 1 (x,0) 2 3 -2 then the dimensions are: length height MTH 173 section 4.7 notes, page 5 28. norman window with perimeter of 30 ft base, perimeter solve for area of window maximize in from a CAS, so critical number is for all so the width should be at the critical number, there is a maximum area, feet. length radius MTH 173 section 4.7 notes, page 6 36.
2r r h R cut R out becomes cup arc = R 2r Relations: In the cone: on the disc: arc length when is in radians between them: the connection: the circumference of the top of the cone is formed from the outside of the disc., so In #36, MTH 173 section 4.7 notes, page 7 Since Now return to area So we are to minimize From a CAS, so then Then The dimensions should be a radius of and a height of is minimized at this critical number MTH 173 section 4.7 notes, page 8 ...
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- Spring '09