**Problem 21**

Find the rectangle of maximum perimeter inscribed in a given circle.

**Solution:**

$x^2 + y^2 = D^2$

$2x + 2y \, y' = 0$

$y' = -x/y$

Perimeter

$P = 2x + 2y$

$dP/dx = 2 + 2y' = 0$

$2 + 2(-x/y) = 0$

$y = x$

The largest rectangle is a square. *answer*

See also the solution using trigonometric function.

**Problem 22**

If the hypotenuse of the right triangle is given, show that the area is maximum when the triangle is isosceles.

**Solution:**

$2x + 2y \, y' = 0$

$y' = -x/y$

Area:

$A = \frac{1}{2} xy$

$dA/db = \frac{1}{2} \, [ \, x y' + y \, ] = 0$

$x y' + y = 0$

$x (-x/y) + y = 0$

$y = x^2 / y$

$y^2 = x^2$

$y = x$

The triangle is an isosceles right triangle. *answer*

**Problem 23**

Find the most economical proportions for a covered box of fixed volume whose base is a rectangle with one side three times as long as the other.

**Solution:**

$V = x \, (3x) \, y

V = 3x^2 \, y$

$0 = 3x^2 \, y' + 6xy$

$y' = -2y/x$

Total Area:

$A_T = 2(3x^2) + 2(3xy) + 2(xy)$

$A_T = 6x^2 + 8xy$

$dA_T / dx = 12x + 8 \, (x y' + y) = 0$

$12x + 8 \, [ \, x (-2y/x) + y \, ] = 0$

$12x + 8 \, [ \, -2y + y \, ] = 0$

$12x = 8y$

$y = \frac{3}{2} x$

Altitude = 3/2 × shorter side of base. *answer*

**Problem 24**

Solve Problem 23 if the box has an open top.

**Solution:**

$V = x \, (3x) \, y$

$V = 3x^2 \, y$

$0 = 3x^2 \, y' + 6xy$

$y' = -2y/x$

Area:

$A = 3x^2 + 2(3xy) + 2(xy)$

$A = 3x^2 + 8xy$

$dA/dx = 6x + 8(x \, y' + y) = 0$

$6x + 8 \, [ \, x (-2y/x) + y \, ] = 0$

$6x + 8 \, [ \, -2y + y \, ] = 0$

$6x = 8y$

$y = \frac{3}{4}x$

Altitude = 3/4 × shorter side of base. *answer*

## Tags:

- 99535 reads