Apple P.

asked • 11/07/21

Math Question (Advanced!!!) Please solve it for me. I am so stuck.

Find the maximum area of a triangle formed in the first quadrant by the x-axis, y-axis, and a tangent line to the graph of f=(x+5)^-2.



Paul M.

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Are you sure you have submitted this question correctly? If you plot the function you have submitted, the graph in the first quadrant is nearly a straight line and the tangent will be very close to f(x).
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11/07/21

Doug C.

Right now I do not have time to solve this using 1st derivative test, etc. BUT the following graph should give some clues as how to proceed. Find the formula for A(x) based on the x and y intercepts of the tangent line, take the first derivative of A(x), set equal to zero, etc. desmos.com/calculator/ogimnbpmes
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11/07/21

1 Expert Answer

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Yefim S. answered • 11/07/21

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Let (X, Y) is tangent point. Then Y = (X + 5)-2; f'(x) = -2(x + 5)-3; f'(X) = -2(X + 5)-3.

Equation of tangent line: y = Y - 2(X + 5)-3(x - X).

Now, we have: y-intercept: x = 0; y = Y + 2X(X+ 5)-3;

x-intercept: y = 0; Y - 2(X+ 5)-3(x - X) = 0; x = Y/2(X + 5)3 + X

Area of triangle: A = 1/2[Y + 2X(X + 5)-3][Y/2(X + 5)3 + X].

We substitude Y = (X + 5)-2. Then Area A = 1/4(X + 5)-3(3X + 5)2. We get area as function of X

dA/dX = -3/4(X+ 5)-4((3X + 5)2 + 3/2(X + 5)-3(3X + 5) = (X + 5)-4(3X + 5)[- 3/4(3X + 5) + 3/2(X + 5)] = 0.

X > 0, - 3X - 5 + 2x + 10 = 0; X = 5

A''(X) = -4(x + 5)-5(3X + 5)(- 3/4X + 15/4) + 3(x + 5)-4(-3/4X + 5) - 3/4(X + 5)-4(3X + 5);

A''(5) = -3/4·10-4·20 < 0. So, we at X = 5 have maximum area: max A + A(5) = 1/4(10)-3(20)2 = 0.1

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