I need a little help with my optimization problems here. Can someone please help me?Thank you in advance ((:
A line is drawn tangent to the curve with equation y=e^x at the point P with coordinates (a,e^a)
a. Determine an equation of the tangent line at the point (a,e^a) and find the coordinates of the point R at which the tangent line crosses the x-axis.
b. Consider the triangle ΔPOR where O is the origin. Express the area of ΔPOR as a function of a.
c. Find the value of a with -1<= a <= 1 for which the area of ΔPOR is a maximum. Justify your answer.
A line is drawn tangent to the curve with equation y=e^x at the point P with coordinates (a,e^a)
a. Determine an equation of the tangent line at the point (a,e^a) and find the coordinates of the point R at which the tangent line crosses the x-axis.
b. Consider the triangle ΔPOR where O is the origin. Express the area of ΔPOR as a function of a.
c. Find the value of a with -1<= a <= 1 for which the area of ΔPOR is a maximum. Justify your answer.
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a. dy/dx = e^x
At (a, e^a), the slope is e^a, so the equation of the tangent line is y - e^a = e^a (x - a).
The x-intercept is found by setting y = 0: -e^a = e^a (x - a).
-e^a = e^a x - ae^a
ae^a - e^a = e^a x
x = a - 1
(a - 1, 0) is the x-intercept.
b. The area of ∆OPR = (1/2)(a - 1)e^a.
c. To maximize the area, A(a), find A'(a) = (1/2)(a - 1)e^a + (1/2)e^a = 0.
(1/2)e^a (a - 1 + 1) = (1/2)e^a a = 0
a = 0
max area = (1/2)(-1)e^0 = 1/2, since area cannot be negative.
At (a, e^a), the slope is e^a, so the equation of the tangent line is y - e^a = e^a (x - a).
The x-intercept is found by setting y = 0: -e^a = e^a (x - a).
-e^a = e^a x - ae^a
ae^a - e^a = e^a x
x = a - 1
(a - 1, 0) is the x-intercept.
b. The area of ∆OPR = (1/2)(a - 1)e^a.
c. To maximize the area, A(a), find A'(a) = (1/2)(a - 1)e^a + (1/2)e^a = 0.
(1/2)e^a (a - 1 + 1) = (1/2)e^a a = 0
a = 0
max area = (1/2)(-1)e^0 = 1/2, since area cannot be negative.