Which quantity must be maximized, and how do price and quantity combine to form that function?

You need to maximize revenue, and revenue is price times quantity. Start by writing

Then find critical points from \(R'(x)=0\), and compare all candidates that lie in \([100,200]\), including endpoints.

At a break-even quantity, what must be true about profit, and therefore about revenue and cost?

Break-even means profit is zero, so revenue equals cost. Set

move all terms to one side, and solve the quadratic equation. Keep only quantities that make sense in context.

How can you use the sign of \(f'(x)\) on intervals to decide whether each critical number is a relative maximum, minimum, or neither?

First factor \(f'(x)=3x^4-12x^2\) to find the critical numbers. Then build a sign chart for \(f'(x)\) on the intervals between those values and classify each point from the sign changes.

For a left-endpoint Riemann sum, which \(x\)-values determine rectangle heights, and how do you find the common width?

Because \([0,8]\) is split into 4 equal pieces, \(\Delta x=(8-0)/4=2\). The left endpoints are \(x=0,2,4,6\), so estimate

What pieces do you need at \(p=3\) before you can evaluate elasticity: the quantity \(x\), the derivative \(dx/dp\), or both?

You need both values. Compute \(x(3)\) from the demand equation and compute \(\left.\dfrac{dx}{dp}\right|_{p=3}\), then substitute into

Finally compare your result to 1 to interpret elastic vs. inelastic demand.

Which formula gives effective annual yield for quarterly compounding, and which formula gives it for continuous compounding?

For Investment A use

and for Investment B use

Convert both to percentages and then compare their difference to \(0.05\%\).

How do marginal cost and total cost relate, and what extra information is needed after integrating?

Integrate \(C'(x)\) term by term to get a general cost function \(C(x)+K\), then use the fixed-cost condition \(C(0)=7000\) to determine the constant and evaluate at \(x=100\).

Does the problem ask for signed net change or geometric area, and how does that affect the sign of your final result?

Start with the definite integral

using antiderivative \(e^{-x}\). Because the curve is below the \(x\)-axis on this interval, take the absolute value at the end to report area.

For area between two curves on an interval, what integrand should you build from the two functions?

Area is

Use a test point in \([0,3]\) to decide which function is on top, then simplify \(g(x)-f(x)\) before matching the choice.

What does each derivative value represent on the graph, and how can you compare them by slope direction and steepness?

At each marked \(x\)-value, sketch or imagine the tangent line. Rank slopes from most negative to most positive by first deciding sign (downward, flat, upward) and then comparing steepness.

What is the quantity being maximized, and how can you write both factors in terms of one variable?

Let \(x\) be the number of extra trees beyond 30. Then total yield is number of trees times yield per tree:

Differentiate \(Y\), solve \(Y'(x)=0\), and then convert back from extra trees to total trees per acre.

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 approximate heights \(f(0), f(2), f(4), f(6)\)

So

Answer: 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 is greater than \(\$26,000\), so the correct choice is E.

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 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 algebraically the integrand is \(-x^2 + 3x\), so the correct setup is

which matches choice D.

Here we reason from the graph. Choose an \(x\)-value, mark the point \((x,f(x))\), and sketch the tangent line at that point. Then compare the steepness (slopes) of those tangent lines.

• \(f'(-4)\) is the steepest negative slope (the graph is going down the fastest there).
• \(f'(-3)\) is negative but near a local minimum, so it is a small negative slope.
• \(f'(2)\) is zero slope (the graph has a horizontal tangent there).
• \(f'(0)\) is a small positive slope (the graph is going up slowly there).
• \(f'(4)\) is the largest positive slope.

So ordering the derivatives, we have:

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.