Some answers to the review problems for the second exam in Math 151:04-06, fall 2003

Although students will work on these review problems on Thursday, I invite e-mail with solutions. Each student first should try the problem assigned to their last name. For example, Jane or John Smith should send me e-mail with the answer to #10. I will post the answers here, and change the red background to green at that time.

Students giving answers will receive my thanks and will be identified here, additionally obtaining notoriety, glory, etc.


Answer to #1

Michael Pandolfo writes:
For part a I got A(1) = 3, ... A'(x) = (23-7(5x^2-3))^(1/2)*(10x) and A'(1) = 30.
b. y-3 = 30(x-1)
c. A(.95) = 1.5
d. Yes A"(1) = -(260/3) The value found in part c is larger than the actual value of A(.95) because A"(x) < 0 so the curve is concave down and the tangent line is above the curve.

Answer to #2

Answer to #3

Answer to #4

Answer to #5

Answer to #6

Answer to #7

Answer to #8

Answer to #9

Answer to #10
Sherman Yang writes:
a. If you input infinity for the top and the bottom, it will be infinity over infinity so you take the l'hopitals. then input infinity to the top and bottom again and it will become 1/infinity which is 0.
b. Just plug in 3 and it is 0/6 so the answer is 0.
c. Separate the equation into two so it will be sinx/(e^x-1)^2 - x/(e^x-1)^2. Since the range of sinx is -1 to 1, you plug in infinity for -1/(e^x-1)^2 and 1/(e^x-1)^2. Both of them will equal 0 so 0

Answer to #11
Sherman Yang writes:
total area = 4x*y and the equation for the amount of fence used in each part is 8x+5y=450. Make this equation equal to y and it becoems y = 90-8/5x. Plug this into the total area equation and it becomes f(x) = 4x* (90- 8/5x) = 360x-(32/5)*x^2. The derivate is f prime x = 360- (64/5)*x. Set it equal to 0 and the C.P. will be 28.125. Plug this into the number line and plug in 2 numbers(lower and higher than 28.125) to the derivate equation and you will see that it is a max. Therefore 28.125 is the highest possible x. Find the y: y= 90-8/5*(28.125) and y =45. The total area is 4*28.125*45 = 5062.5

David Baron writes:
First you set the area (450) equal to 2 times the height (h) plus 5 times the width (w) since the region is divided into four pens. Find w and h. Then since the area equals the height times the width, you substitute what h equals in for w. Find that value, A(H) which equals 90h-(2/5)h^2. Since the slope of the line equals 0, you can find the derivative of A(H). Thus h equals 112.5 m when set to equal 0. Then substitute h back into the original equation and the width comes out to be 45 m. So 45 times 112.5 equals the area of 5062 m^2. To make sure that it's the largest area, you can find the second derivative of A(H). Since the derivative is -(4/5), the curve is concave down which tells us that it's the max value.

Answer to #12

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