Calculus Problems.docx

Calculus Problems 1-3 Derivations 5-7 8. Find the slope of the tangent line to the graph  at the point .    Plugging in  gives    or    Thus, the slope of the tangent line to this graph at the point  is . Since we have a point and the slope of the tangent line, we will use the point-slope form of the line. The equation of the tangent line at  is    or    9. Find the absolute minimum and maximum on  of the function .  It is known that the only place an absolute minimum or maximum can occur on an interval is at one of the endpoints of the interval or at a critical point inside the interval. So start by finding the derivative.    This derivative is defined everywhere so the only possible critical numbers will be where the derivative is equal to 0.   can only be true if  or  but  is always positive so it is never equal to . Thus the only critical point comes when . As stated before, the absolute max and min can only occur at an endpoint or at a critical number. So our only choices for  here are , , or . If these are our only choices, our easiest plan is to just plug in the three values and see which gives the highest value and which gives the lowest value.        Thus, the absolute minimum comes at the point  and the absolute maximum comes at the point .  10. Find the absolute minimum and maximum on  of the function .    This derivative is never undefined. It is equal to 0 at infinitely many values of  since the  function is periodic but on our given interval, it is equal to  only at a finite number of points. This derivative is equal to  either when  or when .  The cosine function is equal to  at any point in the set . So, the derivative will be equal to  whenever  is equal to one of these values. The only values of  inside of our interval for which this is true are  and . Thus, we have five values as candidates for our absolute max and the same five are candidates for our absolute min. Plug in all five values into the original function. The highest will be our absolute max and the lowest will be our absolute min.            Thus, we have an absolute minimum on the interval at the point  and an absolute maximum on the interval at two points, both of which have the same height,  and .  11. Find all local minimums and maximums of the function .  Local minimums and maximums may occur never, once, or many times depending on the function. The derivative of our function is    This derivative is always defined and is equal to  only when . So our only critical point is . To decide if this is a local maximum or minimum let us look at the behavior of the derivative to the left and to the right of it. Since this is the only point where the derivative is ever equal to  it must be true that to the left of  the derivative is always negative or always positive and the same can be said to the right of . If the derivative is positive that tells us that the function is increasing in that region and if the derivative is negative that tells us that the function is decreasing in that region. This information will tell us if the critical point is a local max, a local min, or neither.  Since the behavior is the same for any point left of  and also always the same for any point right of , just pick one point on each side and find the value of the derivative at that x-value.      Since the derivative to the left of  is negative, that tells us that the function is decreasing up to that point and since the derivative is positive to the right, we know the function is increasing on the interval . If the function goes from decreasing to increasing, this must be a local minimum at the point  which is easily verified by looking at a graph.  12. Find all local minimums and maximums of the function  .    A fraction is equal to  only when the numerator is equal to . However,  is always positive so the derivative is never equal to . It is undefined when  but so is the original function. Therefore there are no critical points and no local minimums or maximums.  13. If a farmer wants to make a rectangular pen along a river, so that only 3 sides need to be fenced in, what is the largest area he can fence with 100 feet of fence?    If the total fence to be used is 100 feet, it makes sense that . Thus, our interval for the problem is . Now, the problem asks us to maximize the area. The derivative of a function can help us find the maximum, if one exists, so we need an equation for the area of this rectangle. That is simply        The derivative is equal to  when  so we have one critical point. Thus are absolute maximum area will be found at , , or .        Thus, the greatest area the fence can enclose in the prescribed manner is 1250 square feet when the side adjacent to the river is 50 feet long and the side perpendicular is 25 feet long.  14. If a fence is to be made with three pens, the three connected side-by-side, find the dimensions which give the largest total area if 400 feet of fence are to be used.  This problem describes a large rectangle with length x and height y. However, this rectangle is split into three parts with two extra lengths of fence of length y. We know the total fence to be used is 400 feet, so    From this equation we can see that  which gives us our interval. Now, to maximize the area, we need a function for the area. Since this is a rectangle, the area is simply        Thus, when  we have a critical point and this is our only critical point. Our absolute maximum area, then, must come when , , or .        Thus, our absolute maximum area of 5000 square feet comes when the length is 100 feet and the height is 50 feet.  15. Find the local minimums and maximums of the function .  To find the local minimums and maximums of a function, find the critical points by finding the derivative and finding the points where it is equal to 0 or undefined. If the original function is defined at these points, then these are critical points.    This function is never equal to 0 as the numerator is never equal to 0. It is undefined at . This is our only critical point. However, the square root function is only defined for nonnegative real numbers so  is not defined to the left of . Thus, it can not have a local minimum or maximum at .   does give us the lowest point on the entire square root function as it starts at  and increases there after.  16. Find the slope of the tangent line to the graph  when   Find the derivative of the function.      to  when  is   17. Find the slope of the tangent line to the graph  when   Find the derivative of the function.    to  when  is   18. Find the slope of the tangent line to the graph  when .  This problem is just like past problems except that we are finding the slope of the tangent line at multiple points. First, find the derivative of .    Plugging in our values, we get                Our original graph is that of a parabola. One half of a parabola is increasing. Notice that it not only increases but increases at an increasing rate. You can see that by looking at the graph and now we can see that by looking at the derivative because the derivative gives us the slope of the tangent line or rate of change. Since the derivative is increasing, this indicates that the rate of change is increasing.  19. Find the slope of the tangent line to the graph  at the point .  Here, use implicit differentiation to find . Then plug in our point to find the slope of the tangent line at that specific point. By taking the derivative of both sides we get    Some students are tempted to plug in the point before taking the derivative. However, , for example, is a variable and  is a constant. Thus, plugging in  before taking the derivative will give us a totally different derivative. After taking the derivative, it is okay to plug in the point. We have two options. We can solve for  and then plug in our point or we can plug our point and then solve for . Usually, it is simpler to simply plug in the point. Here, this gives    or    So, if we were to draw the tangent line to this graph, a circle, we would get a horizontal line through the point .  20. Find the slope of the tangent line to the graph  at the point .  Again, use implicit differentiation to find  and then plug in the point  to find the slope of the tangent line at that point. Taking the derivative gives    Plugging in  gives    or    Thus, the slope of the tangent line to this graph at the point  is . Since we have a point and the slope of the tangent line, we will use the point-slope form of the line. The equation of the tangent line at  is    or    21. Find the slope of the tangent line to the graph  when .  Again, take the derivative of the function and plug in  for  and that will give you the slope of the tangent line at .    and    Thus, if we were to draw the tangent line to  at , the line would be a horizontal line with slope . MAXIMA MINIMA 22. Find the rectangle of maximum perimeter inscribed in a given circle. Diameter D is constant (circle is given)   Perimeter   The largest rectangle is a square. 23. If the hypotenuse of the right triangle is given, show that the area is maximum when the triangle is isosceles.   Area:   The triangle is an isosceles right triangle. 24. Find the most economical proportions for a covered box of fixed volume whose base is a rectangle with one side three times as long as the other. Given Volume:   Total Area:   Altitude = 3/2 × shorter side of base. 25. Solve Problem 24 if the box has an open top Given Volume:   Area:   Altitude = 3/4 × shorter side of base. 26. The strength of a rectangular beam is proportional to the breadth and the square of the depth. Find the shape of the largest beam that can be cut from a log of given size. Diameter is given (log of given size), thus D is constant   Strength:   27. The stiffness of a rectangular beam is proportional to the breadth and the cube of the depth. Find the shape of the stiffest beam that can be cut from a log of given size. Diameter is given (log of given size), thus D is constant   Stiffness:   28. Compare for strength and stiffness both edgewise and sidewise thrust, two beams of equal length, 2 inches by 8 inches and the other 4 inches by 6 inches (See Problem 26 and Problem 27 above). Which shape is more often used for floor joist? Why? Strength, S = bd2 Stiffness, k = bd3   right0For 2" × 8": Oriented such that the breadth is 2" S = 8(22) = 32 in3 k = 8(23) = 64 in4 Oriented such that the breadth is 8" S = 2(82) = 128 in3 k = 2(83) = 1024 in4   right0For 4" × 6": Oriented such that the breadth is 6" S = 6(42) = 96 in3 k = 6(43) = 384 in4 Oriented such that the breadth is 4" S = 4(62) = 144 in3 k = 4(63) = 864 in4   2" x 8" is stiffer than 4" x 6" and it is the commonly used size for floor joists. In fact, some local codes required a minimum depth of 8". 29. A box is to be made of a piece of cardboard 9 inches square by cutting equal squares out of the corners and turning up the sides. Find the volume of the largest box that can be made in this way.     Using quadratic formula   Use x = 1.5 inches   Maximum volume: 30. Find the volume of the largest box that can be made by cutting equal squares out of the corners of a piece of cardboard of dimensions 15 inches by 24 inches, and then turning up the sides.     31. A rectangular field of fixed area is to be enclosed and divided into three lots by parallels to one of the sides. What should be the relative dimensions of the field to make the amount of fencing minimum? Area:   Fence:   width = ½ × length 32. Do problem 31 with the words "three lots" replaced by "five lots" Area:   Fence:   33. A rectangular lot is bounded at the back by a river. No fence is needed along the river and there is to be 24-ft opening in front. If the fence along the front costs $1.50 per foot, along the sides $1 per foot, find the dimensions of the largest lot which can be thus fenced in for $300. Total cost:   Area:     Dimensions: 84 ft × 112 ft 34. What should be the shape of a rectangular field of a given area, if it is to be enclosed by the least amount of fencing? Area:   Perimeter:    (a square) 35. A rectangular field of given area is to be fenced off along the bank of a river. If no fence is needed along the river, what is the shape of the rectangle requiring the least amount of fencing? Area:   Perimeter:   width = ½ × length 36. Find two numbers whose sum is a, if the product of one to the square of the other is to be a minimum. Let x and y = the numbers       The numbers are 1/3 a, and 2/3 a. 37. Find two numbers whose sum is a, if the product of one by the cube of the other is to be a maximum. Let x and y the numbers       The numbers are 1/4 a and 3/4 a. 38. Find two numbers whose sum is a, if the product of the square of one by the cube of the other is to be a maximum. Let x and y the numbers           The numbers are 2/5 a and 3/5 a. 39. Find the most economical proportions of a quart can. Volume:   Total area (closed both ends): Diameter = height 40. Find the most economical proportions for a cylindrical cup. Volume:   Area (open one end):   Radius = height 41. Find the most economical proportions for a box with an open top and a square base. Volume:   Area:   Aide of base = 2 × altitude 42. The sum of the length and girth of a container of square cross section is a inches. Find the maximum volume.   Volume     For 2x = 0; x = 0 (meaningless) For a - 6x = 0; x = 1/6 a Use x = 1/6 a     43. Find the proportion of the circular cylinder of largest volume that can be inscribed in a given sphere. From the figure:   Volume of cylinder:   44. In Problem 43 above, find the shape of the circular cylinder if its convex surface area is to be a maximum. Convex surface area (shaded area):   From Solution to Problem 30 above, dh/dd = -d/h 45. Find the dimension of the largest rectangular building that can be placed on a right-triangular lot, facing one of the perpendicular sides. Area:   From the figure:       Dimensions: ½ a × ½ b 46. A lot has the form of a right triangle, with perpendicular sides 60 and 80 feet long. Find the length and width of the largest rectangular building that can be erected, facing the hypotenuse of the triangle. Area:   By similar triangle:       Thus,   Dimensions: 50 ft × 24 ft 47. A page is to contain 24 sq. in. of print. The margins at top and bottom are 1.5 in., at the sides 1 in. Find the most economical dimensions of the page. Print Area:  Page area:     Dimensions: 6 in × 9 in 48. A Norman window consists of a rectangle surmounted by a semicircle. What shape gives the most light for the given perimeter? Given perimeter: Where: Thus,   Light is most if area is maximum: breadth = height 49. Solve Problem 48 above if the semicircle is stained glass admitting only half the normal amount of light. From Solution of Problem 36  Half amount of light is equivalent to half of the area. 50. Two posts, one 8 feet high and the other 12 feet high, stand 15 ft apart. They are to be supported by wires attached to a single stake at ground level. The wires running to the tops of the posts. Where should the stake be placed, to use the least amount of wire? Total length of wire:   For x + 30 = 0; x = -30 (meaningless) For x - 6 = 0; x = 6 (ok) use x = 6 ft   Location of stake is 6 ft from the shorter post.