the legs of a right triangle have lengths a and b satisfying a + b =10. which values of a and b maximize the area of the triangle?

area = (1/2) a b

b = 10 - a

area = (1/2) a (10 -a)

= (1/2) 10 a - a^2

If we were doing algebra we could find the vertex of this parabola but since you said calculus, we will use calculus
dArea/da = 0 when max or min
dArea/da = (1/2)(10 - 2 a)
2 a = 10
a = 5
b = 5

To maximize the area of a right triangle with leg lengths a and b, we need to find the values of a and b that maximize the product of the two leg lengths.

Given that a + b = 10, the sum of the leg lengths, we can express b in terms of a:

b = 10 - a

Now, we can express the area of the triangle using the leg lengths a and b:

Area = (1/2) * a * b = (1/2) * a * (10 - a)

To find the maximum area, we can take the derivative of the area function with respect to a and set it equal to zero:

d(Area)/d(a) = 0

Let's differentiate the area function with respect to a:

d(Area)/d(a) = (1/2) * (10 - 2a) = 0

Now solve for a:

10 - 2a = 0

2a = 10

a = 5

Substituting this value of a back into the equation b = 10 - a, we can find the value of b:

b = 10 - 5 = 5

So, when the leg lengths a and b equal 5, the area of the right triangle is maximized.

To maximize the area of a right triangle with legs of lengths a and b, we need to find the values of a and b that satisfy the given constraint a + b = 10 and yield the largest possible area.

The area of a right triangle is given by the formula: A = (1/2) * base * height.

In this case, the base and height of the triangle are the lengths of its legs, a and b. Therefore, the area can be expressed as: A = (1/2) * a * b.

To find the maximum area, we can use the constraint a + b = 10 to express one of the variables in terms of the other. Since a = 10 - b, we can substitute this expression into the area formula, giving us A = (1/2) * (10 - b) * b.

Now, let's simplify the formula and find the maximum value:

A = (1/2) * (10b - b^2)
= 5b - (b^2 / 2)
= - (b^2 / 2) + 5b

To find the maximum value of A, we need to determine the value of b that maximizes this quadratic equation. This can be achieved by finding the vertex of the parabola represented by the equation.

The x-coordinate of the vertex can be found using the formula: x = -b / (2a), where in this case a = -1/2 and b = 5.

x = -(5) / (2 * (-1/2))
x = -5 / -1
x = 5

Therefore, b = 5 maximizes the area.

Substituting this value back into the constraint a + b = 10, we find:
a + 5 = 10
a = 10 - 5
a = 5

Thus, the values that maximize the area of the triangle are a = 5 and b = 5, with an area of A = (1/2) * 5 * 5 = 12.5 square units.

Is not really okay.