Revenue is

Write it out and find critical points.

To maximize \(R(x)\) on \([100,200]\):

1. Compute \(R'(x)\).
2. Set \(R'(x)=0\) and solve for \(x\).
3. Check whether that critical point lies in \([100,200]\).
4. Evaluate \(R(100)\) and \(R(200)\), and compare with any critical point inside the interval.

Break-even means revenue = cost.

Use

and set up the equation \(R(x) = C(x)\).

1. Start from \[ x(120 - 0.25x) = 0.5x^2 - 375. \]
2. Move everything to one side to get a quadratic \(ax^2 + bx + c = 0\).
3. Use the quadratic formula or calculator program to solve for \(x\).
4. Ignore any negative solution and keep the positive solution that fits one of the ranges.

Critical numbers happen when \(f'(x)=0\).

1. Factorize two times: \[ f'(x) = 3x^4 - 12x^2 = 3x^2(x^2 - 4) = \dots \]
2. Set this equal to zero and find all values of \(x\) that make it zero.

To decide max/min/neither:

1. Make a sign chart for \(f'(x)\) using the critical numbers \(-2,0,2\).
2. Pick test points in
   \((-\infty,-2),\ (-2,0),\ (0,2),\ (2,\infty)\).
3. See where \(f'(x)\) is positive (increasing) or negative (decreasing).
4. Use:
   • increasing → decreasing = max,
   • decreasing → increasing = min,
   • same sign on both sides = neither.

Split \([0,8]\) into 4 equal pieces.

Compute the width:

For a left-endpoint sum:

1. Use \(x=0,2,4,6\) as the bases of the rectangles.
2. From the graph, read \(f(0), f(2), f(4), f(6)\).
3. Compute \[ \text{Area} \approx \Delta x\,[f(0)+f(2)+f(4)+f(6)]. \]

We use the elasticity formula

First differentiate \(x(p)=16-\frac{1}{3}p^2\) to find \(\dfrac{dx}{dp}\).

At \(p=3\):

1. Find \(x(3)\).
2. Find \(\dfrac{dx}{dp}\big|_{p=3}\).
3. Plug into \[ E(3) = -\frac{3}{x(3)}\cdot \frac{dx}{dp}\bigg|_{p=3}. \]
4. Compare \(|E(3)|\) with 1:
   • \(|E(3)|>1\) → elastic,
   • \(|E(3)|<1\) → inelastic.

Use effective-rate formulas:

• For A (compounded \(n=4\) times/year): \[ r_{\text{eff},A} = \left(1+\frac{0.036}{4}\right)^4 - 1. \]
• For B (continuous compounding): \[ r_{\text{eff},B} = e^{0.0355} - 1. \]

1. Approximate both effective rates (as decimals or percentages).
2. Compare them: which one is larger?
3. Check if the difference in rates is greater or less than \(0.0005\) (that’s 0.05%).

Since \(C'(x)\) is the derivative of \(C(x)\), integrate \(C'(x)\) to get \(C(x)\) plus a constant.

Then use the fixed cost \(C(0)=7000\) to find that constant.

After you get

do the following:

1. Plug in \(x=100\).
2. Compute each term carefully (do the powers and coefficients step by step).
3. Add everything and see which interval it falls into among the answer choices.

Start with the definite integral

Find an antiderivative of \(-e^{-x}\).
(Recall: the derivative of \(e^{-x}\) is \(-e^{-x}\).)

1. Evaluate \[ \int_0^1 -e^{-x}dx = \left[e^{-x}\right]_0^1. \]
2. This gives a negative value (since the function is below the x-axis).
3. The geometric area is the positive value, so take the absolute value and then simplify the expression.

Area between curves is

Pick a point in the interval, e.g. \(x=1\), and compare \(f(1)\) and \(g(1)\) to see which is on top.

1. Once you know which is top and bottom, write: \[ \int_0^3 [\text{top} - \text{bottom}]\,dx. \]
2. Compute \[ g(x) - f(x) = (-x+2) - (x^2 - 4x + 2). \]
3. Simplify the integrand and match your result to one of the answer choices.

Remember:

• \(f'(x)\) = slope of the tangent at that point.
• \(m\) is the slope of a secant line from \(-4\) to \(4\).

For each point, decide if the tangent slope is negative, zero, or positive, and whether it’s steep or gentle.

1. Identify where the graph is going down fastest (most negative slope).
2. Identify a smaller negative slope.
3. Find where the tangent looks horizontal (slope \(=0\)).
4. Among positive slopes, see which is largest and which is smaller.
5. Estimate the secant slope \(m\): it should be a “middle” positive slope between some of the tangent values.

Then order them from smallest (most negative) to largest (most positive).

Let \(x\) be the number of extra trees per acre.

• Number of trees: \(30 + x\).
• Yield per tree: starts at 480 and drops by 12 for each extra tree: \[ 480 - 12x. \]

Total yield is “trees × yield per tree.”

1. Write the total yield function: \[ Y(x) = (30 + x)(480 - 12x). \]
2. Expand \(Y(x)\) into standard quadratic form.
3. Differentiate to find \(Y'(x)\).
4. Solve \(Y'(x)=0\) to find the best \(x\).
5. Convert back to the actual number of trees: \(30 + x\).

Step 1: Revenue function

Price:

Revenue = (price)(quantity):

Distribute the \(x\):

We want the maximum of \(R(x)\) on \(100 \le x \le 200\).

Step 2: Find critical point

Derivative:

Set the derivative equal to zero:

Add \(0.5x\) to both sides:

Divide both sides by \(0.5\):

So the critical point is at \(x = 240\), but it is outside the allowed range \([100,200]\).

Step 3: Evaluate endpoints

Because the critical point is outside the interval, the maximum must occur at one of the endpoints.

• At \(x=100\):

  \[ R(100) = 120(100) - 0.25(100)^2 = 12000 - 0.25(10{,}000) = 12000 - 2500 = 9500. \]

• At \(x=200\):

  \[ R(200) = 120(200) - 0.25(200)^2 = 24000 - 0.25(40{,}000) = 24000 - 10{,}000 = 14{,}000. \]

So the absolute maximum revenue on \([100,200]\) is \(\boxed{\$14{,}000}\).

This falls into answer choice D.

We want the break-even point, where

From Problem 1:

Set them equal:

Step 1: Move everything to one side

Add \(0.25x^2\) to both sides and add 375 to both sides:

Now subtract \(120x\) from both sides:

It is easier without decimals. Note that \(0.75 = \frac{3}{4}\). Multiply the entire equation by 4:

So we solve

Step 2: Quadratic formula

For \(ax^2+bx+c=0\), here \(a=3, b=-480, c=-1500\).

Compute inside the square root:

• \( (-480)^2 = 230{,}400\),
• \( 4ac = 4\cdot 3 \cdot (-1500) = 12\cdot (-1500) = -18{,}000\).

So

Thus

\(\sqrt{248{,}400}\) is approximately \(498.4\).

So the two solutions are approximately:

The negative solution does not make sense for a number of tables, so we keep

This is in the interval \(150 < x \le 175\), so the correct answer is C.

We are given:

Step 1: Factor the derivative

Factor out the greatest common factor \(3x^2\):

Now factor the difference of squares:

so

Step 2: Critical numbers

Critical numbers occur when \(f'(x) = 0\):

This happens when

so

Step 3: Sign chart for \(f'(x)\)

We examine intervals:

• \((-\infty,-2)\)
• \((-2,0)\)
• \((0,2)\)
• \((2,\infty)\)

Pick test values:

1. For \(x=-3\) (in \((-\infty,-2)\))

   • \(x^2 > 0\),
   • \(x-2\) is negative,
   • \(x+2\) is negative.

   So the sign of \(f'(x)\) is

   \[ (+)\cdot(-)\cdot(-) = (+), \]

   which means \(f'(x) > 0\) and \(f\) is increasing.

2. For \(x=-1\) (in \((-2,0)\))

   • \(x^2>0\),
   • \(x-2\) is negative,
   • \(x+2\) is positive.

   Signs: \((+)\cdot(-)\cdot(+) = (-)\).
   So \(f'(x) < 0\) and \(f\) is decreasing.

3. For \(x=1\) (in \((0,2)\))

   The signs are the same as for \(-1\), so \(f'(x) < 0\) and \(f\) is still decreasing.

4. For \(x=3\) (in \((2,\infty)\))

   • \(x^2>0\),
   • \(x-2\) is positive,
   • \(x+2\) is positive.

   Signs: \((+)\cdot(+)\cdot(+)=(+)\).
   So \(f'(x) > 0\) and \(f\) is increasing.

Step 4: Classification

Look at how \(f\) changes around each critical point:

• At \(x=-2\):

  To the left, \(f'(x) > 0\) (increasing).
  To the right, \(f'(x) < 0\) (decreasing).
  Increasing followed by decreasing means a relative maximum at \(x=-2\).

• At \(x=0\):

  \(f'(x) < 0\) on both sides (decreasing on left and right).
  Decreasing on both sides means \(x=0\) is neither a max nor a min.

• At \(x=2\):

  To the left, \(f'(x) < 0\) (decreasing).
  To the right, \(f'(x) > 0\) (increasing).
  Decreasing followed by increasing means a relative minimum at \(x=2\).

This matches answer choice B.

We divide \([0,8]\) into 4 equal subintervals.

Step 1: Width \(\Delta x\)

Left endpoints: \(x=0, 2, 4, 6\).

Step 2: Riemann sum formula

Left Riemann sum:

From the graph on the exam, you read the heights \(f(0), f(2), f(4), f(6)\). In this version they are approximately \(3,3,-2,6\), so

This corresponds to answer choice B.

Demand:

Step 1: Compute \(x\) and \(\dfrac{dx}{dp}\)

Differentiate with respect to \(p\):

At \(p=3\):

and

Step 2: Elasticity formula

So

Because \(E(3) < 1\), demand is inelastic at price $3.

Answer: B.

We compare effective annual yields.

Step 1: Investment A (3.6% quarterly)

Nominal rate \(r_A = 0.036\), compounded \(n=4\) times per year.

Effective rate:

Compute:

Thus

Step 2: Investment B (3.55% continuous)

Nominal rate \(r_B = 0.0355\).

Effective rate:

Using a calculator,

So

Step 3: Compare

Difference:

Investment A has the better yield, but only by 0.03%, which is less than 0.05%.

Answer: C.

We are given the marginal cost:

and fixed cost \(C(0)=7000\).

We want \(C(100)\).

Step 1: Integrate to find \(C(x)\)

Integrate term by term:

1. \(\displaystyle \int 0.000009x^2\,dx = 0.000009 \cdot \frac{x^3}{3} = 0.000003x^3.\)

2. \(\displaystyle \int (-0.08x)\,dx = -0.08 \cdot \frac{x^2}{2} = -0.04x^2.\)

3. \(\displaystyle \int 200\,dx = 200x.\)

So

Step 2: Use fixed cost to find \(K\)

Using \(C(0)=7000\):

so

Thus

Step 3: Evaluate at \(x=100\)

Compute each term:

• \(0.000003x^3\) with \(x=100\):

  \(100^3 = 1{,}000{,}000\), so
  

  \[ 0.000003 \cdot 1{,}000{,}000 = 3. \]

• \(-0.04x^2\) with \(x=100\):

  \(100^2 = 10{,}000\), so
  

  \[ -0.04 \cdot 10{,}000 = -400. \]

• \(200x\) with \(x=100\):

  \[ 200\cdot 100 = 20{,}000. \]

Now sum:

So \(C(100) = \$26{,}603\).

This falls in the range \(\$24,000 \le C(100) \le \$26{,}000\), so answer D.

We want the area bounded by the curve \(y=-e^{-x}\) and the \(x\)-axis from \(x=0\) to \(x=1\).

Note: the curve lies below the \(x\)-axis (since \(-e^{-x} < 0\)), but area is positive.

Step 1: Compute the definite integral

The signed area (integral of the function) is

Antiderivative:

(since the derivative of \(e^{-x}\) is \(-e^{-x}\)).

Evaluate from 0 to 1:

This value is negative (because \(e^{-1}<1\)).

Step 2: Turn signed area into geometric area

Area between the curve and the x-axis is the absolute value:

This equals \(-e^{-1}+1\) (just reversing the order of subtraction).

So the correct expression is

which is answer A.

We have

We want the area between them from \(x=0\) to \(x=3\).

Step 1: Decide which function is on top

Evaluate at a convenient point, say \(x=1\):

• \(f(1) = 1^2 - 4(1) + 2 = 1 - 4 + 2 = -1\).
• \(g(1) = -1 + 2 = 1\).

So at \(x=1\), \(g(x)\) is above \(f(x)\).
The graph on the exam also shows \(g\) above \(f\) on \([0,3]\).

Thus the top function is \(g(x)\) and the bottom function is \(f(x)\).

Step 2: Set up area integral

Compute \(g(x) - f(x)\):

Distribute the minus sign carefully:

So the integrand is \(-x^2 + 3x\).

In the multiple-choice list from the original paper exam, the intended correct option corresponds to an equivalent area expression and is labeled D. For the purposes of this practice page, you can think of the correct setup as

and then choose the option that matches this integrand.

We reason from the graph.

• \(f'(-4)\) is the steepest negative slope (the graph is going down the fastest there).
• \(f'(-3)\) is negative but not as steep (a smaller negative slope).
• \(f'(2)\) is the slope at a local extremum, so it is zero (horizontal tangent).
• \(f'(0)\) is a small positive slope (the graph is going up slowly there).
• \(f'(4)\) is the largest positive slope (the graph is rising the fastest there).
• \(m\), the secant slope from \(-4\) to \(4\), is a moderate positive slope, between the smaller and larger positive derivatives.

Putting them in order from least (most negative) to greatest (most positive):

which matches answer choice B on the original exam.

We model total yield (pounds per acre).

Let \(x\) be the number of extra trees per acre, beyond 30.

Then:

• Number of trees: \(30+x\).
• Yield per tree: starting at 480 lb, decreasing by 12 lb for each extra tree: \[ \text{yield per tree} = 480 - 12x. \]

Step 1: Total yield function

Expand:

First multiply \(30\) by each term:

Then multiply \(x\) by each term:

Add them:

So

This is a downward-opening parabola (the coefficient of \(x^2\) is negative), so it has a maximum.

Step 2: Take derivative and find critical point

Set the derivative equal to zero:

Subtract 120 from both sides:

Divide by \(-24\):

So the critical point is at \(x=5\).

Step 3: Confirm maximum

Second derivative:

Since \(Y''(x) = -24 < 0\), the graph is concave down, so \(x=5\) gives a maximum.

Step 4: Interpret in terms of trees

Recall: \(x\) is the number of extra trees beyond 30.

So the optimal number of trees is:

Thus, to maximize total yield, the farmer should plant 35 trees per acre.

Use the slope formula

with the two points, then plug \(m\) and one point into point–slope form:

If \(C(x)\) is linear, then marginal cost is just the slope of the line:

Compute that and interpret it as “dollars per cup.”

For a quadratic \(ax^2 + bx + c\), the vertex has

Plug in \(a=3\), \(b=-14\).

For rational functions where numerator and denominator have the same degree, the horizontal asymptote is

Profit = Revenue − Cost:

Just subtract the expressions carefully.

Vertical asymptotes for a rational function occur where the denominator is zero (and the numerator is nonzero). Set the denominator equal to zero and solve for \(x\).

Use the continuous compounding formula

Compute \(A\) and then interest = \(A - P\).

Use

with \(r_{\text{nom}} = 0.082\) and \(n=4\). Then convert the decimal to a percent.

For continuous compounding,

Set \(e^r - 1 = 0.061\) and solve for \(r\) using natural logs:

Look at the left-hand limit and right-hand limit as \(x\) approaches 1 on the graph. If they agree and hit a specific value, that’s the limit; if they differ, the limit does not exist.

The derivative at \(x=2\) is the slope of the tangent line at that point. Draw (mentally) a line that just touches the graph at \(x=2\) and estimate its slope.

Use the quotient rule:

with \(u = 7x + 3\), \(v = 2x - 5\).

Remember: \(P'(x)\) is the rate of change of profit w.r.t. \(x\). So \(P'(20)=10\) means at sales level \(x=20\), profit is changing by about $10 per additional T-shirt.

Apply the power rule term by term:

and constant terms differentiate to zero.

Marginal revenue is

Differentiate \(R(q)\) and then plug in \(q=3400\).

Use the product rule:

Then evaluate at \(x=3\) using the given numbers.

Use chain rule with \(u = 2x-1\):

So you’ll get a fraction with denominator \(2x-1\).

First simplify:

Then differentiate term by term using the power rule.

Use chain rule:

Differentiate the exponent, then multiply by the original exponential.

Use the quotient rule:

Then plug in \(x=3\).

1. Write attendance \(N(p)\) as a linear function of \(p\) using the starting point and slope (change in people per $1).
2. Revenue: \[ R(p) = (\text{ticket revenue per person} + 14)\cdot N(p) = (p + 14) N(p). \]
3. Expand to get a quadratic in \(p\), then maximize (vertex or derivative).

1. Find the point: compute \(g(0)\).
2. Find the slope: \(g'(x) = 3e^{3x}\), so \(g'(0)=3\).
3. Use point–slope form: \[ y - g(0) = g'(0)(x - 0). \]

1. Compute the point on the curve: \(g(5)\).
2. Find the derivative \(g'(x)\) and evaluate \(g'(5)\).
3. Use \(y - g(5) = g'(5)(x-5)\).

Average cost:

1. Simplify \(\bar C(x)\).
2. Differentiate \(\bar C(x)\).
3. Find critical points in \((0,10]\) and compare values at those points and at \(x=10\).

Critical numbers are \(x\)-values where \(f'(x)=0\) (horizontal tangent) or where \(f'(x)\) does not exist, as long as \(f\) is defined there. Look for peaks, valleys, and sharp corners.

“\(f\) is increasing” where the graph is going up as you move left to right. Identify intervals in \(x\) where the curve rises.

1. Solve \(f'(x)=0\) to get critical points.
2. Make a small sign chart for \(f'(x)\) around those points.
3. Use sign changes (+ to − or − to +) to determine max vs min.

Relative maxima/minima of \(f\) occur where \(f'\) changes sign:

• \(f'\) goes + to − → local max in \(f\).
• \(f'\) goes − to + → local min in \(f\). Look where the derivative graph crosses the \(x\)-axis.

Use

1. Differentiate \(q(p)\) to get \(dq/dp\).
2. Evaluate \(q(40)\) and \(\frac{dq}{dp}\big|_{p=40}\).
3. Plug into the formula.

Again

Here \(dq/dp\) requires the chain rule on \(e^{-0.02p}\).

Use the power rule for integrals:

Apply it term by term and remember the constant of integration \(C\).

Again, integrate each term separately using the power rule, and add \(+C\) at the end.

1. Find the width: \[ \Delta x = \frac{4 - (-2)}{3}. \]
2. Left endpoints: \(x_0=-2\), \(x_1=-2+\Delta x\), \(x_2=-2+2\Delta x\).
3. Approximate: \[ \text{Area} \approx \Delta x\,[f(x_0)+f(x_1)+f(x_2)]. \]

1. Integrate \(C'(x)\) with respect to \(x\) to get \(C(x) + K\).
2. Use \(C(0) = 70\) (fixed cost) to solve for the constant \(K\).

1. Find intersection points by solving \(x^2 + 1 = x^3\).
2. Decide which function is on top on that interval.
3. Area: \[ \int_a^b [\text{top} - \text{bottom}]\,dx. \]

Try substitution: let

Then \(du = (2x+8)\,dx\). The integrand becomes \(e^u\,du\).

So the definite integral is

You can simplify using log properties.

Use

Factor out 45000, integrate \(e^{-0.031t}\), and evaluate from 0 to 8.

1. Find the equilibrium price \(p_0 = D(50)\).
2. Use \[ \text{CS} = \int_0^{q_0} D(q)\,dq - p_0 q_0. \] Compute the integral, then subtract the rectangle.

1. Find equilibrium price \(p_0 = S(80)\).
2. Producers’ surplus: \[ \text{PS} = p_0 x_0 - \int_0^{x_0} S(x)\,dx. \] Integral first, then subtract from the big rectangle.