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will be V ( x) = x ( 12 2x) ( 12 2x) = 144x 48x 2 + 4x 3 Our job is to maximize this function V Now V ( x) = 144 96x + 12x 2 We may solve the quadratic equation 144 96x + 12x 2 = 0 to find the critical points for this problem Using the quadratic formula, we find that x = 2 and x = 6 are the critical points for the problem Now V ( x) = 96 + 24x Since V ( 2) = 48 < 0, we conclude that x = 2 is a local maximum for the problem In fact we can sketch a graph of V ( x) using ideas from calculus and see that x = 2 is a global maximum We conclude that if squares of side 2 are cut from the sheet of cardboard then a box of maximum volume will result Observe in passing that if squares of side 6 are cut from the sheet then (there will be no cardboard left!) the resulting box will have zero volume This value for x gives a minimum for the problem

The bottom of the ladder is pulled fiom the wall at 15m/s What is the rate at which the top of the ladder is going down when the bottom is 3 m fiom the wall

6 7 8 9 10

A rectangular garden is to be constructed against the side of a garage The gardener has 100 feet of fencing, and will construct a three-sided fence; the side of the garage will form the fourth side What dimensions will give the garden of greatest area

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Look at Figure 314 Let x denote the side of the garden that is perpendicular to the side of the garage Then the resulting garden has width x feet and length 100 2x feet The area of the garden is A( x) = x ( 100 2x) = 100x 2x 2 We calculate A ( x) = 100 4x and find that the only critical point for the problem is x = 25 Since A ( x) = 4 for all x, we determine that x = 25 is a local maximum By inspection, we see that the graph of A is a downward-opening parabola So x = 25 must also be the global maximum that we seek The optimal dimensions for the garden are width = 25 ft length = 50 ft

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Solution: Help to visualize the problem by sketching a ladder leaning against a wall with the bottom being pulled out fiom the wall at dx/dt = 15m/s (Fig 6-1)

YOU TRY IT Find the right circular cylinder of greatest volume that can be contained in a sphere of radius 1

36 37 38 39 40

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Let x be one of the two numbers Then the other is 60 x Their product is P ( x) = x ( 60 x) = 60x x 2 Thus P is the quantity that we wish to maximize Calculating the derivative, we find that P ( x) = 60 2x Thus the only critical point for the problem is x = 30 Since P ( x) 2, we find that x = 30 is a local maximum Since the graph of P is a downward-opening parabola, we can in fact be sure that x = 30 is a global maximum We conclude that the two numbers that add to 60 and maximize the product are 30 and 30

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YOU TRY IT A rectangular box is to be placed in the first quadrant {( x, y) :

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The question, written in mathematical language, is: "What is dy/dt when x = 3 m and &/dt = 15 m/s "

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x 0, y 0} in such a way that one side lies on the positive x-axis and one side lies on the positive y-axis The box is to lie below the line y = 2x + 5 Give the dimensions of such a box having greatest possible area

44655 46489 47813 48613 48881

127

41 42 43 44 45

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