{"id":1143,"date":"2026-01-18T13:56:27","date_gmt":"2026-01-18T03:56:27","guid":{"rendered":"https:\/\/archive4ones.com\/2ndstudy\/?p=1143"},"modified":"2026-01-18T13:56:27","modified_gmt":"2026-01-18T03:56:27","slug":"year11-math-3-1-3","status":"publish","type":"post","link":"https:\/\/archive4ones.com\/2ndstudy\/?p=1143","title":{"rendered":"Year11-MATH-3-1-3"},"content":{"rendered":"\n

Chapter 3: Applications of the Derivative<\/h1>\n\n\n\n

In Chapter 2, we learned how to find the derivative (f<\/mi>\u2032<\/mo><\/msup>(<\/mo>x<\/mi>)<\/mo><\/mrow>f'(x)<\/annotation><\/semantics><\/math>) to determine the gradient of a curve. Now, we apply that skill to solve real-world problems. By finding where a gradient is zero, we can identify the highest and lowest points of a function\u2014a process called Optimization<\/strong>.<\/p>\n\n\n\n
\n\n\n\n

3.1 Stationary Points<\/h2>\n\n\n\n

A stationary point<\/strong> occurs where the gradient of the function is zero (f<\/mi>\u2032<\/mo><\/msup>(<\/mo>x<\/mi>)<\/mo>=<\/mo>0<\/mn><\/mrow>f'(x) = 0<\/annotation><\/semantics><\/math>). At these points, the tangent to the curve is perfectly horizontal.<\/p>\n\n\n\n

There are three main types of stationary points:<\/p>\n\n\n\n

    \n
  1. Local Maximum:<\/strong> The “peak” of a hill. The gradient changes from positive to negative.<\/li>\n\n\n\n
  2. Local Minimum:<\/strong> The “bottom” of a valley. The gradient changes from negative to positive.<\/li>\n\n\n\n
  3. Stationary Point of Inflection:<\/strong> The curve flattens out momentarily but then continues in the same direction.<\/li>\n<\/ol>\n\n\n\n
    \n\n\n\n

    3.2 Determining the Nature of a Point<\/h2>\n\n\n\n

    To find out if a stationary point is a maximum or minimum, we use a Sign Table<\/strong> to check the gradient just before and just after the point.<\/p>\n\n\n\n

    x<\/mi>x<\/annotation><\/semantics><\/math><\/td>a<\/mi>\u2212<\/mo>\u03f5<\/mi><\/mrow>a\u2212\u03f5<\/annotation><\/semantics><\/math> (Just before)<\/strong><\/td>a<\/mi>a<\/annotation><\/semantics><\/math> (The point)<\/strong><\/td>a<\/mi>+<\/mo>\u03f5<\/mi><\/mrow>a+\u03f5<\/annotation><\/semantics><\/math> (Just after)<\/strong><\/td><\/tr><\/thead>
    f<\/mi>\u2032<\/mo><\/msup>(<\/mo>x<\/mi>)<\/mo><\/mrow>f'(x)<\/annotation><\/semantics><\/math><\/td>Positive (+)<\/td>0<\/mn>0<\/annotation><\/semantics><\/math><\/td>Negative (-)<\/td><\/tr>
    Shape<\/strong><\/td>\/<\/td>\u2014<\/td>\\<\/td><\/tr>
    Conclusion<\/strong><\/td><\/td>Local Maximum<\/strong><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
    \n\n\n\n

    3.3 Increasing and Decreasing Functions<\/h2>\n\n\n\n
      \n
    • A function is increasing<\/strong> on an interval if f<\/mi>\u2032<\/mo><\/msup>(<\/mo>x<\/mi>)<\/mo>><\/mo>0<\/mn><\/mrow>f'(x) > 0<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
    • A function is decreasing<\/strong> on an interval if f<\/mi>\u2032<\/mo><\/msup>(<\/mo>x<\/mi>)<\/mo><<\/mo>0<\/mn><\/mrow>f'(x) < 0<\/annotation><\/semantics><\/math>.<\/li>\n<\/ul>\n\n\n\n
      \n

      Worked Example 1: Finding Stationary Points<\/strong> Find the coordinates and nature of the stationary points for f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>3<\/mn><\/msup>\u2212<\/mo>3<\/mn>x<\/mi><\/mrow>f(x) = x^3 – 3x<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n
        \n
      1. Differentiate:<\/strong> f<\/mi>\u2032<\/mo><\/msup>(<\/mo>x<\/mi>)<\/mo>=<\/mo>3<\/mn>x<\/mi>2<\/mn><\/msup>\u2212<\/mo>3<\/mn><\/mrow>f'(x) = 3x^2 – 3<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
      2. Set to Zero:<\/strong> 3<\/mn>x<\/mi>2<\/mn><\/msup>\u2212<\/mo>3<\/mn>=<\/mo>0<\/mn>\u2192<\/mo>3<\/mn>x<\/mi>2<\/mn><\/msup>=<\/mo>3<\/mn>\u2192<\/mo>x<\/mi>2<\/mn><\/msup>=<\/mo>1<\/mn>\u2192<\/mo>x<\/mi>=<\/mo>1<\/mn>,<\/mo>\u2212<\/mo>1<\/mn><\/mrow>3x^2 – 3 = 0 \\rightarrow 3x^2 = 3 \\rightarrow x^2 = 1 \\rightarrow x = 1, -1<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
      3. Find <\/strong>y<\/mi>y<\/annotation><\/semantics><\/math>-coordinates:<\/strong> > * f<\/mi>(<\/mo>1<\/mn>)<\/mo>=<\/mo>(<\/mo>1<\/mn>)<\/mo>3<\/mn><\/msup>\u2212<\/mo>3<\/mn>(<\/mo>1<\/mn>)<\/mo>=<\/mo>\u2212<\/mo>2<\/mn><\/mrow>f(1) = (1)^3 – 3(1) = -2<\/annotation><\/semantics><\/math>. Point is (<\/mo>1<\/mn>,<\/mo>\u2212<\/mo>2<\/mn>)<\/mo><\/mrow>(1, -2)<\/annotation><\/semantics><\/math>.\n
          \n
        • f<\/mi>(<\/mo>\u2212<\/mo>1<\/mn>)<\/mo>=<\/mo>(<\/mo>\u2212<\/mo>1<\/mn>)<\/mo>3<\/mn><\/msup>\u2212<\/mo>3<\/mn>(<\/mo>\u2212<\/mo>1<\/mn>)<\/mo>=<\/mo>2<\/mn><\/mrow>f(-1) = (-1)^3 – 3(-1) = 2<\/annotation><\/semantics><\/math>. Point is (<\/mo>\u2212<\/mo>1<\/mn>,<\/mo>2<\/mn>)<\/mo><\/mrow>(-1, 2)<\/annotation><\/semantics><\/math>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
        • Test Nature (for <\/strong>x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math>):<\/strong>\n
            \n
          • Check x<\/mi>=<\/mo>0<\/mn><\/mrow>x=0<\/annotation><\/semantics><\/math>: f<\/mi>\u2032<\/mo><\/msup>(<\/mo>0<\/mn>)<\/mo>=<\/mo>\u2212<\/mo>3<\/mn><\/mrow>f'(0) = -3<\/annotation><\/semantics><\/math> (Negative).<\/li>\n\n\n\n
          • Check x<\/mi>=<\/mo>2<\/mn><\/mrow>x=2<\/annotation><\/semantics><\/math>: f<\/mi>\u2032<\/mo><\/msup>(<\/mo>2<\/mn>)<\/mo>=<\/mo>3<\/mn>(<\/mo>4<\/mn>)<\/mo>\u2212<\/mo>3<\/mn>=<\/mo>9<\/mn><\/mrow>f'(2) = 3(4)-3 = 9<\/annotation><\/semantics><\/math> (Positive).<\/li>\n\n\n\n
          • Since it goes Negative \u2192<\/mo>\\to<\/annotation><\/semantics><\/math> Zero \u2192<\/mo>\\to<\/annotation><\/semantics><\/math> Positive, (<\/mo>1<\/mn>,<\/mo>\u2212<\/mo>2<\/mn>)<\/mo><\/mrow>(1, -2)<\/annotation><\/semantics><\/math> is a Local Minimum.<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/blockquote>\n\n\n\n
            \n\n\n\n

            3.4 Optimization in Practice<\/h2>\n\n\n\n

            Optimization is the process of finding the “best” value (e.g., maximum area or minimum cost).<\/p>\n\n\n\n

            The Strategy:<\/strong><\/p>\n\n\n\n

              \n
            1. Create an equation for the variable you want to optimize (e.g., Area A<\/mi>A<\/annotation><\/semantics><\/math>).<\/li>\n\n\n\n
            2. Ensure the equation has only one<\/strong> independent variable (you may need to substitute).<\/li>\n\n\n\n
            3. Differentiate the equation.<\/li>\n\n\n\n
            4. Set the derivative to zero and solve.<\/li>\n<\/ol>\n\n\n\n
              \n

              Worked Example 2: Maximizing Area<\/strong> <\/p>\n\n\n\n

              A farmer has 40m of fencing to create a rectangular paddock against a straight wall (only 3 sides need fencing). Find the dimensions that maximize the area.<\/p>\n\n\n\n

                \n
              1. Let the sides perpendicular to the wall be <\/strong>x<\/mi>x<\/annotation><\/semantics><\/math>. Let the side parallel be y<\/mi>y<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
              2. Constraint:<\/strong> 2<\/mn>x<\/mi>+<\/mo>y<\/mi>=<\/mo>40<\/mn>\u2192<\/mo>y<\/mi>=<\/mo>40<\/mn>\u2212<\/mo>2<\/mn>x<\/mi><\/mrow>2x + y = 40 \\rightarrow y = 40 – 2x<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
              3. Area Formula:<\/strong> A<\/mi>=<\/mo>x<\/mi>\u00d7<\/mo>y<\/mi>=<\/mo>x<\/mi>(<\/mo>40<\/mn>\u2212<\/mo>2<\/mn>x<\/mi>)<\/mo>=<\/mo>40<\/mn>x<\/mi>\u2212<\/mo>2<\/mn>x<\/mi>2<\/mn><\/msup><\/mrow>A = x \\times y = x(40 – 2x) = 40x – 2x^2<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
              4. Differentiate:<\/strong> d<\/mi>A<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac>=<\/mo>40<\/mn>\u2212<\/mo>4<\/mn>x<\/mi><\/mrow>\\frac{dA}{dx} = 40 – 4x<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
              5. Set to Zero:<\/strong> 40<\/mn>\u2212<\/mo>4<\/mn>x<\/mi>=<\/mo>0<\/mn>\u2192<\/mo>4<\/mn>x<\/mi>=<\/mo>40<\/mn>\u2192<\/mo>x<\/mi>=<\/mo>10<\/mn><\/mrow>40 – 4x = 0 \\rightarrow 4x = 40 \\rightarrow x = 10<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
              6. Find <\/strong>y<\/mi>y<\/annotation><\/semantics><\/math>:<\/strong> y<\/mi>=<\/mo>40<\/mn>\u2212<\/mo>2<\/mn>(<\/mo>10<\/mn>)<\/mo>=<\/mo>20<\/mn><\/mrow>y = 40 – 2(10) = 20<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
              7. Answer:<\/strong> Dimensions are 10m by 20m<\/strong> for a maximum area of 200m\u00b2<\/strong>.<\/li>\n<\/ol>\n<\/blockquote>\n\n\n\n
                \n\n\n\n

                3.5 Practice Problems<\/h2>\n\n\n\n

                Part A: Stationary Points<\/h3>\n\n\n\n
                  \n
                1. Find the stationary points of f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>2<\/mn><\/msup>\u2212<\/mo>6<\/mn>x<\/mi>+<\/mo>8<\/mn><\/mrow>f(x) = x^2 – 6x + 8<\/annotation><\/semantics><\/math> and determine their nature.<\/li>\n\n\n\n
                2. For y<\/mi>=<\/mo>2<\/mn>x<\/mi>3<\/mn><\/msup>\u2212<\/mo>9<\/mn>x<\/mi>2<\/mn><\/msup>+<\/mo>12<\/mn>x<\/mi><\/mrow>y = 2x^3 – 9x^2 + 12x<\/annotation><\/semantics><\/math>, find the coordinates of the local maximum and local minimum.<\/li>\n<\/ol>\n\n\n\n

                  Part B: Intervals<\/h3>\n\n\n\n
                    \n
                  1. Determine the interval for which f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>x<\/mi><\/mrow>f(x) = x^2 – 4x<\/annotation><\/semantics><\/math> is a decreasing function.<\/li>\n<\/ol>\n\n\n\n

                    Part C: Optimization<\/h3>\n\n\n\n
                      \n
                    1. The height of a projectile is given by h<\/mi>(<\/mo>t<\/mi>)<\/mo>=<\/mo>30<\/mn>t<\/mi>\u2212<\/mo>5<\/mn>t<\/mi>2<\/mn><\/msup><\/mrow>h(t) = 30t – 5t^2<\/annotation><\/semantics><\/math>, where t<\/mi>t<\/annotation><\/semantics><\/math> is time in seconds. Find the maximum height reached.<\/li>\n\n\n\n
                    2. Challenge:<\/strong> A box with a square base and no top must have a volume of 32<\/mn> cm<\/mtext>3<\/mn><\/msup><\/mrow>32 \\text{ cm}^3<\/annotation><\/semantics><\/math>. Find the side length of the base (x<\/mi>x<\/annotation><\/semantics><\/math>) that minimizes the surface area. (Hint: V<\/mi>=<\/mo>x<\/mi>2<\/mn><\/msup>h<\/mi><\/mrow>V = x^2h<\/annotation><\/semantics><\/math>, S<\/mi>A<\/mi>=<\/mo>x<\/mi>2<\/mn><\/msup>+<\/mo>4<\/mn>x<\/mi>h<\/mi><\/mrow>SA = x^2 + 4xh<\/annotation><\/semantics><\/math>).<\/li>\n<\/ol>\n\n\n\n
                      \n\n\n\n

                      Solutions (Summary)<\/h3>\n\n\n\n
                        \n
                      • 1.<\/strong> f<\/mi>\u2032<\/mo><\/msup>(<\/mo>x<\/mi>)<\/mo>=<\/mo>2<\/mn>x<\/mi>\u2212<\/mo>6<\/mn>\u2192<\/mo>x<\/mi>=<\/mo>3<\/mn><\/mrow>f'(x) = 2x – 6 \\rightarrow x = 3<\/annotation><\/semantics><\/math>. Point is (<\/mo>3<\/mn>,<\/mo>\u2212<\/mo>1<\/mn>)<\/mo><\/mrow>(3, -1)<\/annotation><\/semantics><\/math>. It is a Local Minimum<\/strong>.<\/li>\n\n\n\n
                      • 2.<\/strong> y<\/mi>\u2032<\/mo><\/msup>=<\/mo>6<\/mn>x<\/mi>2<\/mn><\/msup>\u2212<\/mo>18<\/mn>x<\/mi>+<\/mo>12<\/mn>=<\/mo>6<\/mn>(<\/mo>x<\/mi>\u2212<\/mo>1<\/mn>)<\/mo>(<\/mo>x<\/mi>\u2212<\/mo>2<\/mn>)<\/mo><\/mrow>y’ = 6x^2 – 18x + 12 = 6(x-1)(x-2)<\/annotation><\/semantics><\/math>. Max at (<\/mo>1<\/mn>,<\/mo>5<\/mn>)<\/mo><\/mrow>(1, 5)<\/annotation><\/semantics><\/math>, Min at (<\/mo>2<\/mn>,<\/mo>4<\/mn>)<\/mo><\/mrow>(2, 4)<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
                      • 3.<\/strong> f<\/mi>\u2032<\/mo><\/msup>(<\/mo>x<\/mi>)<\/mo>=<\/mo>2<\/mn>x<\/mi>\u2212<\/mo>4<\/mn><\/mrow>f'(x) = 2x – 4<\/annotation><\/semantics><\/math>. Decreasing when 2<\/mn>x<\/mi>\u2212<\/mo>4<\/mn><<\/mo>0<\/mn>\u2192<\/mo>x<\/mi><<\/mo>2<\/mn><\/mrow>2x – 4 < 0 \\rightarrow x < 2<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
                      • 4.<\/strong> h<\/mi>\u2032<\/mo><\/msup>(<\/mo>t<\/mi>)<\/mo>=<\/mo>30<\/mn>\u2212<\/mo>10<\/mn>t<\/mi>=<\/mo>0<\/mn>\u2192<\/mo>t<\/mi>=<\/mo>3<\/mn><\/mrow>h'(t) = 30 – 10t = 0 \\rightarrow t = 3<\/annotation><\/semantics><\/math> seconds. h<\/mi>(<\/mo>3<\/mn>)<\/mo>=<\/mo>45<\/mn>m<\/mtext><\/mrow>h(3) = 45 \\text{m}<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
                      • 5.<\/strong> h<\/mi>=<\/mo>32<\/mn>x<\/mi>2<\/mn><\/msup><\/mfrac><\/mrow>h = \\frac{32}{x^2}<\/annotation><\/semantics><\/math>. S<\/mi>A<\/mi>=<\/mo>x<\/mi>2<\/mn><\/msup>+<\/mo>4<\/mn>x<\/mi>(<\/mo>32<\/mn>x<\/mi>2<\/mn><\/msup><\/mfrac>)<\/mo>=<\/mo>x<\/mi>2<\/mn><\/msup>+<\/mo>128<\/mn>x<\/mi><\/mfrac><\/mrow>SA = x^2 + 4x(\\frac{32}{x^2}) = x^2 + \\frac{128}{x}<\/annotation><\/semantics><\/math>. S<\/mi>A<\/mi>\u2032<\/mo><\/msup>=<\/mo>2<\/mn>x<\/mi>\u2212<\/mo>128<\/mn>x<\/mi>2<\/mn><\/msup><\/mfrac>=<\/mo>0<\/mn>\u2192<\/mo>x<\/mi>3<\/mn><\/msup>=<\/mo>64<\/mn>\u2192<\/mo>x<\/mi>=<\/mo>4<\/mn> cm<\/mtext><\/mrow>SA’ = 2x – \\frac{128}{x^2} = 0 \\rightarrow x^3 = 64 \\rightarrow x = 4 \\text{ cm}<\/annotation><\/semantics><\/math>.<\/li>\n<\/ul>\n\n\n\n
                        \n\n\n\n

                        <\/p>\n","protected":false},"excerpt":{"rendered":"

                        Chapter 3: Applications of the Derivative In Chapter 2, we learned how to find the derivative (f\u2032(x)f'(x)) to determine the gradient of a curve. Now, we apply that skill to solve real-world problems. By finding where a gradient is zero, we can identify the highest and lowest points of a function\u2014a process called Optimization. 3.1 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[8],"tags":[],"class_list":["post-1143","post","type-post","status-publish","format-standard","hentry","category-math"],"_links":{"self":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/1143","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1143"}],"version-history":[{"count":2,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/1143\/revisions"}],"predecessor-version":[{"id":1146,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/1143\/revisions\/1146"}],"wp:attachment":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1143"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1143"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1143"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}