# 4.6: Homework- Optimization

- Page ID
- 88662

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- Samantha has some whiskey at a party, and (being a science and math geek) estimates her blood alcohol content (BAC) follows the function:
\(BAC(t) = \frac{0.25 t}{e^t},\)

where \(t\) is measured in hours after her first drink. Graph this function, and determine the following using a derivative:

- How quickly is her BAC increasing (or decreasing) 15 minutes after her first drink?
\(BAC'(t) = \frac{0.25 - 0.25t}{e^t}\), (grams per dL per hour).ans
- How quickly is her BAC increasing (or decreasing) 1 hour after her first drink?
\(0\) changeans
- How quickly is her BAC increasing (or decreasing) 2 hours after her first drink?
\(\approx 0.034\) grams per dL per hourans

- How quickly is her BAC increasing (or decreasing) 15 minutes after her first drink?
- Graph each function over the given interval. Use calculus to determine the location of all global and local mins and maxes.
- \(f(x) = -x^2 + 5x - 2\) on the interval \([0, 5]\).
Local and Global Mins: \((0, -2), (5, -2)\), Local and global max: \((2.5, 4.25)\)ans
- \(f(x) = x^2 - 6x + 10\) on the interval \([2, 6]\).
Local min: \((2, 2)\), global and local min: \((3, 1)\), local and global max: \((6, 10)\)ans
- \(f(x) = x^3 - 6x^2 + 11x - 6\) on the interval \([0, 3]\).
Local and global maximum at \((1.42, 0.38 )\), Local min: \((2.58,-0.38 )\), local and global minimum: \((0, -6)\), local maximum: \((3, -2)\)ans
- \(f(x) = x^3-5 x^2+8 x-4\) on the interval \([0, 2.5]\).
Local max at \(x = 0.75\), local min at \(x = 2\), global max at \(x = 2.5\), global min at \(x = 0\).ans
- \(f(x) = x^4 - 16x\) on the interval \([0, 3]\).
Local max at \(x = 0\), global min at \(x = \sqrt[3]{4}\), global max at \(x = 3\)ans
- \(f(x) = \frac{x^2}{x + 1}\) on the interval \([-3, 3]\).
Local min at \(x = -3\), local max at \(x = -2\), local min at \(x = 0\), local max at \(x = 3\)ans

- \(f(x) = -x^2 + 5x - 2\) on the interval \([0, 5]\).
- Using a chemotherapy drug on a petri-dish of cancer cells, it is found that \(P(x)\) percent more of the cancer cells are killed using \(x\) milligrams of drug per square centimeter than healthy cells, where \(x\) ranges from \(0\) to \(4\). It is thought
\(P(x) = x^3-8 x^2+16 x\)

For what value of \(x\) is \(P(x)\) maximized?

ans - Bananas as we know them may be doomed! Suppose the fungus Tropical Race 4 mentioned in the article is killing off bananas on an island in Jamaica. The number of viable banana farms starts at \(16000\), with \(800\) being forced to close per year. But new banana farms are according to the function \(20e^{0.3t}\) with new varieties immune to the fungus (\(t\) measured in years). So the total number of viable banana farms is
\(V(t) = 16000- 800t + 20e^{0.3t}\)

on the interval \([0, 20]\). At what point is the number of banana farms minimized? What is the number of viable banana farms at this point?

This function is minimized at \(t \approx 16.3\) with the number of banana farms at \(5619\)ans - The area of a rectangle is length times width. A farmer needs to build a pig pen against the side of the barn using \(20\) meters of fence. What is the maximum amount of area he can enclose?
- Watch the KhanAcademy videos on maximizing the area of a box:

Optimizing Box Volume Graphically and

Optimizing Box Volume Analytically - An open-topped box is formed by removing the square corners of sidelength \(x\) off of a \(40\) in by \(80\) in piece of cardboard, and folding each side up. What value of \(x\) maximizes the volume of the box?
\(x \approx 8.45\)ans
- The height and radius of a cone together add to \(5\) inches. What value of the radius maximizes the volume? The volume is given by \(V = \frac{1}{3} \pi r^2 h\).
inans