The sum of the base and the height of a triangle is 2626 cm. find the dimensions for which the area is a maximum.

Mathematics

1 answer:

Lelu [443]3 years ago

7 0

Let the height of the triangle be $h$ and the base be $b$ .

Given that $h+b=26$ and the area of the triangle is $A=\frac{bh}{2}$ . Or

$A(h)=\frac{(26-h)h}{2}$ .

When the area is maximum, $\frac{dA}{dh} =0$ .

or,

$\frac{d\frac{(26-h)h}{2} }{dh} =0$ .

$\frac{d(26-h)h }{dh} =0$

$2h-26=0$

$h=13$

The dimensions of the triangle with maximum area is

$h=13 cm, b=13 cm$

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Help me out please and thank you

faust18 [17]

<h3>Answer: 1</h3>

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Explanation:

We're given the height in relation to base BC, so we need to find the length of this base. This is the same as finding the distance from B to C.

Turn to the distance formula

$d = \text{Distance from B to C}\\\\d = \sqrt{(x_1-x_2)^2+(y_1-y^2)^2}\\\\d = \sqrt{(-2-(-3))^2+(-1-(-2))^2}\\\\d = \sqrt{(-2+3)^2+(-1+2)^2}\\\\d = \sqrt{(1)^2+(1)^2}\\\\d = \sqrt{1+1}\\\\d = \sqrt{2}\\\\$

Coincidentally, the base and height are the same. This won't always be the case.

Now we can find the area of the triangle

$A = \text{Area of triangle}\\\\A = \frac{1}{2}*\text{Base}*\text{Height}\\\\A = \frac{1}{2}*\sqrt{2}*\sqrt{2}\\\\A = \frac{1}{2}*\sqrt{2*2}\\\\A = \frac{1}{2}*\sqrt{4}\\\\A = \frac{1}{2}*2\\\\A = 1\\\\$