Corrected practice final exam: differential calculus (201-103)

The form is indeterminate, $\dfrac{0}{0}$. Factor: $2x^{2}-18=2\left(x^{2}-9\right)=2\left(x-3\right)\left(x+3\right)$, so for $x\neq 3$ the expression equals $2\left(x+3\right)$. The limit is $2\times 6=12$.

Form $\dfrac{0}{0}$ again. Multiply by the conjugate: $\dfrac{\sqrt{x}-2}{x-4}\times \dfrac{\sqrt{x}+2}{\sqrt{x}+2}=\dfrac{x-4}{\left(x-4\right)\left(\sqrt{x}+2\right)}=\dfrac{1}{\sqrt{x}+2}$. The limit is $\dfrac{1}{2+2}=\dfrac{1}{4}$.

For the first, the numerator and denominator have equal degrees: the limit is the ratio of the leading coefficients, $\dfrac{3}{2}$. For the second, the denominator has the higher degree: the limit is 0. The general rule comes in three cases: numerator of higher degree, the limit is infinite; equal degrees, ratio of the leading coefficients; denominator of higher degree, limit zero.

We need $k=\displaystyle\lim_{x\to 3}f(x)=12$. The function extended by $f(3)=12$ is then continuous, and the discontinuity is called REMOVABLE: it came only from a single missing point in the graph.

Three conditions: $f(a)$ must EXIST; the limit at $a$ must EXIST; and the two must be EQUAL. With $k=10$, the first two are satisfied, $f(3)=10$ and the limit is 12, but the third fails: $f(3)\neq \lim_{x\to 3}f(x)$. The graph therefore has an isolated point at 10 while the curve passes through 12.

Product rule: $f'(x)=6x\left(x-4\right)+\left(3x^{2}+1\right)\left(1\right)=6x^{2}-24x+3x^{2}+1=9x^{2}-24x+1$. Check: expanding, $f(x)=3x^{3}-12x^{2}+x-4$, so $f'(x)=9x^{2}-24x+1$. The two methods agree.

Quotient rule: $g'(x)=\dfrac{2\left(x^{2}+1\right)-\left(2x+5\right)\left(2x\right)}{\left(x^{2}+1\right)^{2}}=\dfrac{2x^{2}+2-4x^{2}-10x}{\left(x^{2}+1\right)^{2}}=\dfrac{-2x^{2}-10x+2}{\left(x^{2}+1\right)^{2}}$. We can factor out $-2$: $g'(x)=\dfrac{-2\left(x^{2}+5x-1\right)}{\left(x^{2}+1\right)^{2}}$.

Chain rule: $h'(x)=5\left(x^{2}-3x\right)^{4}\left(2x-3\right)$. The classic error is to forget the factor $2x-3$, that is, the derivative of the inside function.

$f'(2)=9\times 4-48+1=36-48+1=-11$. And $f(2)=\left(12+1\right)\left(-2\right)=-26$. Tangent: $y=-26-11\left(x-2\right)=-11x-4$.

$f$ is decreasing where $f'(x)<0$, that is, $9x^{2}-24x+1<0$. Roots: $x=\dfrac{24\pm \sqrt{576-36}}{18}=\dfrac{24\pm \sqrt{540}}{18}$, with $\sqrt{540}\approx 23.238$, so $x_{1}\approx 0.0423$ and $x_{2}\approx 2.624$. The parabola $9x^{2}-24x+1$ opens upward, so it is negative BETWEEN its roots: $f$ is decreasing on the interval $\left(0.042,\ 2.624\right)$ and increasing elsewhere.

$P'(t)=500\times 0.04\,e^{0.04t}=20\,e^{0.04t}$. At $t=10$: $P'(10)=20\,e^{0.4}\approx 20\times 1.49182\approx 29.84$ people per year. The population is therefore growing by about 30 people per year at that moment, against 20 per year at the start: the growth itself is speeding up.

The relative rate is $\dfrac{P'(t)}{P(t)}=\dfrac{20\,e^{0.04t}}{500\,e^{0.04t}}=\dfrac{20}{500}=0.04$, that is 4% per year, independent of $t$. This is the property that defines exponential growth: the ABSOLUTE change increases, the RELATIVE change stays constant.

Solve $500\,e^{0.04t}=1{,}000$, so $e^{0.04t}=2$ and $0.04t=\ln 2$, hence $t=\dfrac{\ln 2}{0.04}=\dfrac{0.693147}{0.04}\approx 17.33$ years. This agrees with the rule of 70: $\dfrac{70}{4}=17.5$ years, the difference coming from rounding $69.3$ up to 70.

$R'(x)=\dfrac{200}{x+1}$. At $x=9$: $R'(9)=\dfrac{200}{10}=20$. The next thousand dollars would therefore bring in about 20 thousand dollars of extra revenue, against 200 for the very first thousand: the return on spending collapses as spending rises.

$R''(x)=-\dfrac{200}{\left(x+1\right)^{2}}<0$ for all $x>-1$: the function is CONCAVE DOWN everywhere, which is the mathematical translation of diminishing returns. Each additional dollar brings in less than the one before, without ever bringing in a negative amount, since $R'>0$. The campaign is therefore always worth pursuing in terms of gross revenue, but $R'(x)$ must be compared with the cost of the next thousand, which is 1, to know how far to go: here $\dfrac{200}{x+1}=1$ gives $x=199$ thousand dollars.

$f'(x)=3x^{2}-12x+9=3\left(x^{2}-4x+3\right)=3\left(x-1\right)\left(x-3\right)$. The critical points are therefore $x=1$ and $x=3$.

$f'$ is an upward-opening parabola, so it is positive before 1, negative between 1 and 3, positive after 3. The function INCREASES on $\left(-\infty,\ 1\right]$, DECREASES on $\left[1,\ 3\right]$ and increases on $\left[3,\ +\infty\right)$. At $x=1$, $f$ goes from increasing to decreasing: LOCAL MAXIMUM, with $f(1)=1-6+9+2=6$. At $x=3$, the reverse: LOCAL MINIMUM, with $f(3)=27-54+27+2=2$.

$f''(x)=6x-12$, which vanishes at $x=2$ and changes sign: this is an inflection point, with value $f(2)=8-24+18+2=4$. Notice that it lies exactly halfway between the two extrema, a general property of cubics.

$f''(x)<0$ for $x<2$: the function is CONCAVE DOWN on $\left(-\infty,\ 2\right)$. $f''(x)>0$ for $x>2$: it is CONCAVE UP on $\left(2,\ +\infty\right)$. The change of concavity can be seen on the graph near $x=2$.

On a closed interval, the values at the critical points AND at the endpoints must be compared. $f(0)=2$, $f(1)=6$, $f(3)=2$, $f(4)=64-96+36+2=6$. The ABSOLUTE MAXIMUM is 6, attained at $x=1$ AND at $x=4$; the ABSOLUTE MINIMUM is 2, attained at $x=0$ AND at $x=3$. Forgetting the endpoints is the most common error in this type of question, and here it would have missed half of the answers.

Volume constraint: $x^{2}h=48$. Cost: the base is $x^{2}$ square metres at 6 dollars, and the four walls are each $xh$ square metres at 4 dollars, so $C=6x^{2}+16xh$.

From the constraint, $h=\dfrac{48}{x^{2}}$, so $C(x)=6x^{2}+16x\times \dfrac{48}{x^{2}}=6x^{2}+\dfrac{768}{x}$, for $x>0$.

$C'(x)=12x-\dfrac{768}{x^{2}}$. Set it to zero: $12x=\dfrac{768}{x^{2}}$, so $12x^{3}=768$ and $x^{3}=64$, hence $x=4$ metres. Check: $C''(x)=12+\dfrac{1{,}536}{x^{3}}>0$ for all $x>0$, the function is concave up and the critical point is indeed a MINIMUM. Height: $h=\dfrac{48}{16}=3$ metres.

$C(4)=6\times 16+\dfrac{768}{4}=96+192=288$ dollars. The base costs 96 dollars and the walls 192 dollars, exactly DOUBLE. This ratio is no accident: at the optimum of a cost of the form $ax^{2}+\dfrac{b}{x}$, the term in $\dfrac{1}{x}$ is always twice the term in $x^{2}$, as setting the derivative to zero shows.

The cost becomes $C=6x^{2}+24xh=6x^{2}+\dfrac{1{,}152}{x}$, so $C'(x)=12x-\dfrac{1{,}152}{x^{2}}$, which is zero when $x^{3}=96$, that is $x=\sqrt[3]{96}\approx 4.579$ metres and $h=\dfrac{48}{20.96}\approx 2.29$ metres. The container becomes WIDER and SHORTER: when the walls cost more, the optimum reduces their area by flattening the container. The minimum cost rises to about $6\times 20.96+\dfrac{1{,}152}{4.579}\approx 125.8+251.6=377.4$ dollars.

Revenue: $R(q)=\left(90-0.25q\right)q=90q-0.25q^{2}$. Profit: $P(q)=90q-0.25q^{2}-0.25q^{2}-18q-1{,}250=72q-0.5q^{2}-1{,}250$. Marginal cost: $C'(q)=0.5q+18$. Marginal revenue: $R'(q)=90-0.5q$.

$P'(q)=72-q$, zero at $q=72$, with $P''=-1<0$: maximum. Marginal check: $R'(72)=90-36=54$ and $C'(72)=36+18=54$. The two coincide.

Price: $p=90-0.25\times 72=90-18=72$ dollars. Profit: $P(72)=72\times 72-0.5\times 5{,}184-1{,}250=5{,}184-2{,}592-1{,}250=1{,}342$ dollars. Total cost: $C(72)=0.25\times 5{,}184+1{,}296+1{,}250=1{,}296+1{,}296+1{,}250=3{,}842$, so average cost $\dfrac{3{,}842}{72}\approx 53.36$ dollars. Unit margin: $72-53.36=18.64$ dollars, which gives back $18.64\times 72\approx 1{,}342$ dollars.

$\overline{C}(q)=0.25q+18+\dfrac{1{,}250}{q}$, with derivative $0.25-\dfrac{1{,}250}{q^{2}}$, zero when $q^{2}=5{,}000$, that is $q\approx 70.71$ units. The average cost there is $0.25\times 70.71+18+\dfrac{1{,}250}{70.71}\approx 17.68+18+17.68=53.36$ dollars. The quantity of minimum average cost, 70.7, is very close to that of maximum profit, 72, but it differs from it: the two optima do not coincide in general.

$C(73)=0.25\times 5{,}329+18\times 73+1{,}250=1{,}332.25+1{,}314+1{,}250=3{,}896.25$. $C(72)=3{,}842$. Exact cost of the 73rd unit: $3{,}896.25-3{,}842=54.25$ dollars, against the 54 dollars given by the marginal cost. The gap, 25 cents, is exactly the coefficient of the quadratic term of the cost, 0.25, as in any quadratic cost model.

$\dfrac{dq}{dp}=-4$ and, at $p=72$, $q=360-288=72$. So $E=-4\times \dfrac{72}{72}=-4$.

The absolute value is 4, well above 1: demand is ELASTIC at the optimal price, in line with the general result that a profit-maximizing firm always sits in the elastic range. This can be checked with the formula $R'(q)=p\left(1+\dfrac{1}{E}\right)=72\left(1-0.25\right)=54$, which gives back the marginal revenue computed earlier.

The elasticity is $-1$ when $-4\dfrac{p}{360-4p}=-1$, so $4p=360-4p$, that is $p=45$ dollars. The quantity is then $360-180=180$ units and the revenue $45\times 180=8{,}100$ dollars, which is indeed the maximum revenue, as $R'(q)=90-0.5q=0$ at $q=180$ confirms.

Revenue is maximal at 45 dollars, profit at 72 dollars: the gap is 27 dollars, or 60% more. The profit-maximizing price is ALWAYS higher than the revenue-maximizing one as soon as marginal cost is positive, since the condition $R'=C'>0$ is reached before $R'$ vanishes. A firm that aims for sales volume therefore sells too much, at too low a price.

At 60 dollars: $q=360-240=120$ units. Profit: $P(120)=72\times 120-0.5\times 14{,}400-1{,}250=8{,}640-7{,}200-1{,}250=190$ dollars, against 1,342 dollars at the optimum. The firm sells 48 more units and loses 1,152 dollars of profit. Revenue, for its part, rises, from $72\times 72=5{,}184$ to $60\times 120=7{,}200$ dollars: this is exactly the situation of a manager who celebrates a rise in sales while ruining profitability.

$\dfrac{dC}{dt}=C'(q)\dfrac{dq}{dt}=\left(0.5\times 60+18\right)\times 4=\left(30+18\right)\times 4=48\times 4=192$ dollars per month.

$P'(q)=72-q$. At $q=60$: $\dfrac{dP}{dt}=\left(72-60\right)\times 4=12\times 4=48$ dollars per month, the profit is rising. At $q=90$: $\dfrac{dP}{dt}=\left(72-90\right)\times 4=-18\times 4=-72$ dollars per month, the profit is falling. The sign of the rate follows that of $P'$, so the position of $q$ relative to 72.

The profit stops increasing when $P'(q)=0$, that is $q=72$. Starting from $q=60$ and gaining 4 units per month, it takes $\dfrac{72-60}{4}=3$ months to get there. Beyond that, each additional month loses money, and the right decision is to stop increasing production.

$L=0.4P$, so $\dfrac{dL}{dt}=0.4\dfrac{dP}{dt}=0.4\times 20\,e^{0.04t}$. At $t=10$: $0.4\times 29.84\approx 11.93$ homes per year. The constant factor 0.4 simply carries over to the derivative: this is the simplest case of the chain rule, that of a composition with a linear function.

$\overline{C}(q)=0.25q+18+\dfrac{1{,}250}{q}$, so $\overline{C}'(q)=0.25-\dfrac{1{,}250}{q^{2}}$. At $q=60$: $0.25-\dfrac{1{,}250}{3{,}600}=0.25-0.347\overline{2}\approx -0.0972$. Then $\dfrac{d\overline{C}}{dt}=-0.0972\times 4\approx -0.389$ dollar per month: the average cost is still FALLING, which is consistent since 60 is below the quantity of minimum average cost, 70.7. It will rise beyond that point.

FALSE. A derivative equal to zero is NECESSARY but not sufficient: $f'$ must CHANGE SIGN. The function $f(x)=x^{3}$ has $f'(0)=0$ without an extremum, the derivative staying positive on both sides; the point is a plateau, or an inflection point with a horizontal tangent. Correct statement: $x=1$ is a critical point; the sign of $f'$ on both sides, or the second derivative, must be examined to conclude.

FALSE. On a CLOSED interval, the absolute maximum can occur at an ENDPOINT, where the derivative does not vanish. On the cubic of exercise 4, $f(4)=6$ equals the local maximum $f(1)=6$, and $f(0)=2$ equals the local minimum $f(3)=2$: forgetting the endpoints would have missed half of the answers. Correct statement: the values at the critical points AND at both endpoints must be compared.

FALSE, the derivative of the inside function is missing. By the chain rule, the derivative is $5\left(x^{2}-3x\right)^{4}\left(2x-3\right)$. Correct statement: differentiate the outside function keeping the inside intact, then MULTIPLY by the derivative of the inside.