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Extremum problems with exponential and logarithmic functions

Extremum problems with exponential and logarithmic functions

The steps of a solution

An extremum problem is solved in the same steps as in grade 11; only the functions are new:

  • choose a variable, and write its interval from the problem;
  • write the quantity to make largest or smallest as a function of the variable;
  • differentiate, and solve f′(x)=0\displaystyle {f'(x)}={0};
  • check with the sign of f′\displaystyle f' (or of f′′\displaystyle f'') that the point is the maximum or minimum wanted;
  • on a closed interval, compare with the values at the ends too;
  • answer in words, with units.
A medicine’s concentration in the blood

The concentration of a medicine in the blood, t\displaystyle t hours after it was taken, is C(t)=20te−0.5t\displaystyle {C(t)}={20te^{-0.5t}} milligrams per litre (t≥0\displaystyle t\geq0).
When is the concentration highest, and what is it?

The example below finds the maximum of C\displaystyle C for t≥0\displaystyle t\geq0:

  • C′(t)=20e−0.5t+20t⋅(−0.5)e−0.5t=20e−0.5t(1−0.5t)\displaystyle {C'(t)}={20e^{-0.5t}+20t\cdot(-0.5)e^{-0.5t}}={20e^{-0.5t}(1-0.5t)}.
  • e−0.5t>0\displaystyle e^{-0.5t}>0, so C′(t)=0\displaystyle {C'(t)}={0} only where 1−0.5t=0\displaystyle {1-0.5t}={0}, that is t=2\displaystyle {t}={2}.
  • Where 0≤t<2\displaystyle 0\leq t<2, C′(t)>0\displaystyle C'(t)>0 – the concentration rises; where t>2\displaystyle t>2, C′(t)<0\displaystyle C'(t)<0 – it falls. So t=2\displaystyle {t}={2} is an absolute maximum.
  • C(2)=40e−1=40e≈14.7\displaystyle {C(2)}={40e^{-1}}={\frac{40}{e}\approx14.7}.

The highest concentration, about 14.7 milligrams per litre, comes two hours after the medicine is taken.

C(x)=20xe−0.5x\displaystyle C(x)=20xe^{-0.5x}
M=(2,14.715)\displaystyle M=(2,14.715)
A rectangle under the graph of e−x2\displaystyle e^{-x^2}

Two vertices of a rectangle are on the x-axis, and the other two on the graph of y=e−x2\displaystyle {y}={e^{-x^2}}.
What is the largest area of such a rectangle?

The example below names the vertices (±x,0)\displaystyle (\pm x,0) and (±x,e−x2)\displaystyle \left(\pm x,e^{-x^2}\right), x>0\displaystyle x>0 – the graph is symmetric about the y-axis:

f(x)=e−x2\displaystyle f(x)=e^{-x^{2}}
A=(−0.707,0)\displaystyle A=(-0.707,0)
B=(0.707,0)\displaystyle B=(0.707,0)
C=(0.707,0.607)\displaystyle C=(0.707,0.607)
D=(−0.707,0.607)\displaystyle D=(-0.707,0.607)
  • The base 2x\displaystyle 2x, the height e−x2\displaystyle e^{-x^2}, the area S(x)=2xe−x2\displaystyle {S(x)}={2xe^{-x^2}}.
  • S′(x)=2e−x2+2x⋅(−2x)e−x2=2e−x2(1−2x2)\displaystyle {S'(x)}={2e^{-x^2}+2x\cdot(-2x)e^{-x^2}}={2e^{-x^2}(1-2x^2)}.
  • S′(x)=0\displaystyle {S'(x)}={0} where x2=12\displaystyle {x^2}={\frac{1}{2}}, that is x=12≈0.71\displaystyle {x}={\frac{1}{\sqrt{2}}\approx0.71} (x>0\displaystyle x>0).
  • S′>0\displaystyle S'>0 left of the point and S′<0\displaystyle S'<0 right of it: a maximum.
  • S(12)=2e−12≈0.86\displaystyle {S\left(\frac{1}{\sqrt{2}}\right)}={\sqrt{2}e^{-\frac{1}{2}}\approx0.86}.
The triangle a tangent cuts from the axes

At the point (t,e−t)\displaystyle (t,e^{-t}), t≥0\displaystyle t\geq0, a tangent is drawn to the graph of y=e−x\displaystyle {y}={e^{-x}}.
The tangent and the axes make a triangle.
What is the largest area of such a triangle?

The example below writes the area in terms of t\displaystyle t:

  • The slope of the tangent: (e−x)′=−e−x\displaystyle {\left(e^{-x}\right)'}={-e^{-x}}, and at the point −e−t\displaystyle -e^{-t}. The tangent: y−e−t=−e−t(x−t)\displaystyle {y-e^{-t}}={-e^{-t}(x-t)}.
  • The y-axis (x=0\displaystyle {x}={0}): y=e−t(1+t)\displaystyle {y}={e^{-t}(1+t)}. The x-axis (y=0\displaystyle {y}={0}): x=t+1\displaystyle {x}={t+1}.
  • The area: A(t)=(t+1)⋅e−t(t+1)2=(t+1)2e−t2\displaystyle {A(t)}={\frac{(t+1)\cdot e^{-t}(t+1)}{2}}={\frac{(t+1)^2e^{-t}}{2}}.
f(x)=e−x\displaystyle f(x)=e^{-x}
y=−e−1x+2e−1\displaystyle y=-e^{-1}x+2e^{-1}
P=(1,0.368)\displaystyle P=(1,0.368)
  • A′(t)=2(t+1)e−t−(t+1)2e−t2=e−t(t+1)(1−t)2\displaystyle {A'(t)}={\frac{2(t+1)e^{-t}-(t+1)^2e^{-t}}{2}}={\frac{e^{-t}(t+1)(1-t)}{2}}, and A′(t)=0\displaystyle {A'(t)}={0} where t=1\displaystyle {t}={1} (t≥0\displaystyle t\geq0).
  • A′>0\displaystyle A'>0 where 0≤t<1\displaystyle 0\leq t<1 and A′<0\displaystyle A'<0 where t>1\displaystyle t>1: a maximum.
  • A(1)=2e−1=2e≈0.74\displaystyle {A(1)}={2e^{-1}}={\frac{2}{e}\approx0.74}. On the graph: the tangent at the point (1,e−1)\displaystyle (1,e^{-1}).
The absolute extremum on a closed interval

On a closed interval, the largest and the smallest value of a function are at an extremum point or at an end of the interval.

The example below finds the extreme values of f(x)=ln⁡xx\displaystyle {f(x)}={\frac{\ln x}{x}} on the interval 1≤x≤e2\displaystyle 1\leq x\leq e^2:

  • f′(x)=1−ln⁡xx2=0\displaystyle {f'(x)}={\frac{1-\ln x}{x^2}}={0} where x=e\displaystyle {x}={e}, inside the interval.
  • The values: f(1)=0\displaystyle {f(1)}={0}, f(e)=1e≈0.37\displaystyle {f(e)}={\frac{1}{e}\approx0.37}, f(e2)=2e2≈0.27\displaystyle {f(e^2)}={\frac{2}{e^2}\approx0.27}.
  • The absolute maximum 1e\displaystyle \frac{1}{e}, at x=e\displaystyle {x}={e}; the absolute minimum 0\displaystyle 0, at the end x=1\displaystyle {x}={1}.
Exercise 1

A shop sells a product at a price of x\displaystyle x shekels.
The shop’s daily revenue from the product is R(x)=100xe−0.1x\displaystyle {R(x)}={100xe^{-0.1x}} shekels.
At what price is the daily revenue highest?

a

Find R′(x)\displaystyle R'(x).

R(x)=100xe−0.1x\displaystyle R(x)=100xe^{-0.1x}

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

    b

    Solve the equation R′(x)=0\displaystyle {R'(x)}={0}.

    c

    Calculate the highest revenue, rounded to the agora.

    R(x)=100xe−0.1x\displaystyle R(x)=100xe^{-0.1x}

      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

      Exercise 2

      For every x>0\displaystyle x>0 there is a point on the graph of y=ln⁡x\displaystyle {y}={\ln x} and a point on the line y=x\displaystyle {y}={x}, one above the other.
      The vertical distance between the two points is d(x)=x−ln⁡x\displaystyle {d(x)}={x-\ln x}.
      Find the smallest vertical distance.

      f(x)=ln⁡x\displaystyle f(x)=\ln x
      g(x)=x\displaystyle g(x)=x
      a

      Find d′(x)\displaystyle d'(x).

      d(x)=x−ln⁡x\displaystyle d(x)=x-\ln x

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

        b

        Solve the equation d′(x)=0\displaystyle {d'(x)}={0}.

        c

        Calculate the smallest distance.

        d(x)=x−ln⁡x\displaystyle d(x)=x-\ln x

          / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

          / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

          Exercise 3

          One vertex of a rectangle is at the origin, and two of its sides are on the axes.
          The vertex opposite the origin is on the graph of y=3−ln⁡x\displaystyle {y}={3-\ln x}, for 0<x<e3\displaystyle 0<x<e^3.
          The area of the rectangle is S(x)=x(3−ln⁡x)\displaystyle {S(x)}={x(3-\ln x)}.

          f(x)=3−ln⁡x\displaystyle f(x)=3-\ln x
          a

          Find S′(x)\displaystyle S'(x).

          S(x)=x(3−ln⁡x)\displaystyle S(x)=x(3-\ln x)

            / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

            / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

            b

            Solve the equation S′(x)=0\displaystyle {S'(x)}={0}.

            c

            Calculate the largest area.

            S(x)=x(3−ln⁡x)\displaystyle S(x)=x(3-\ln x)

              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

              d

              Explain with the sign of S′\displaystyle S' that this is indeed a maximum.

              Exercise 4

              The function f(x)=x2e−x\displaystyle {f(x)}={x^2e^{-x}} is given, on the interval −1≤x≤3\displaystyle -1\leq x\leq3.
              Find the largest and the smallest value of the function on the interval.

              a

              Find f′(x)\displaystyle f'(x).

              f(x)=x2e−x\displaystyle f(x)=x^{2}e^{-x}

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                b

                Solve the equation f′(x)=0\displaystyle {f'(x)}={0}.

                c

                Calculate the values of the function at the ends of the interval and at the extremum point other than x=0\displaystyle {x}={0} (f(0)=0\displaystyle {f(0)}={0}).

                f(x)=x2e−x\displaystyle f(x)=x^{2}e^{-x}

                  / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                  / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                  f(x)=x2e−x\displaystyle f(x)=x^{2}e^{-x}

                    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                    f(x)=x2e−x\displaystyle f(x)=x^{2}e^{-x}

                      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                      d

                      Write the largest and the smallest value, and where each one is reached.