2007 Test 2 solution

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10 Apr 2012
MAT 137Y, 2007–2008 Winter Session – Solutions to Term Test 2
(15%) 1. A farmer with 1500 feet of fencing wants to enclose a rectangular area and then divide it into four
pens (fenced enclosures) with fencing parallel to one side of the rectangle. Find the largest possible
total area of the four pens; be sure to verify that the area is indeed maximized.
Draw a diagram. Let xbe the length of the side of the field with the five parallel fences. The amount
of fencing used is therefore 5x+2y=1500. We are asked to maximize the rectangular area, that is,
A=xy =x(750 5
2x), where xmust be between 0 and 300. To do this, we find A0:
A(x) = 750x5
2x2=A0(x) = 750 5x=0=x=150.
At the endpoints, A=0, so the critical point must yield the maximum area, hence the largest possible
area is 150 ·375 =56250 square feet.
(10%) (i) Find the values of aand bsuch that lim
Let Lbe the limit above. The numerator and denominator both go to zero, so by L’Hˆ
Since the denominator of the new limit goes to zero, the limit exists only if the numerator goes
to zero. Therefore, b=2. Applying L’Hˆ
opital’s Rule again twice gives us
x04sin 2x+6ax
x08cos 2x+6a
which is true when a=4
3. Hence a=4
3and b=2.
(10%) (ii) The length of a rectangle is increasing at a rate of 7 cm/sec and the width is decreasing at 2
cm/sec. When the length is 20 cm and the width is 6 cm, determine whether the area of the
rectangle is increasing or decreasing, and find the rate at which the area is changing.
Let lbe the length and wbe the width. The area of the rectangle is A=lw, so
dt =l·dw
dt +w·dl
dt .
When l=20 and w=6, we have dl
dt =7 and dw
dt =2, so
dt =20 ·(2) + 6·7=2,
so the area is increasing at a rate of 2 square cm per second.
3. Let f(x) = 2xx.
(5%) (a) Find the domain of f(x)and, if any, the intercepts and asymptotes.
The domain is all x0. Intercepts occur at (0,0)and when
so (4,0)is a y-intercept. As the function is defined for all values of x0 and lim
x0f(x) = 0 and
f(x) = , there are no asymptotes.
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