The perimeter of the rectangle is 32 and the area of the rectangle is 60. Find the larger side.

Let one side of the rectangle be x and the other side be y. The perimeter of the rectangle is equal to the sum of the lengths of the sides of the rectangle (P = 2 (a + b)), i.e. 2 (x + y) or 32. The area of ​​a rectangle is equal to the product of its sides (S = ab), ie xy or 60. Let’s compose a system of equations and solve it.

2 (x + y) = 32; xy = 60;

x + y = 32: 2; xy = 60;

x + y = 16; xy = 60; – express from the first equation y through x;

y = 16 – x; – substitute the expression (16 – x) in the second equation instead of y;

x (16 – x) = 60;

16x – x ^ 2 = 60;

x ^ 2 – 16x + 60 = 0;

D = b ^ 2 – 4ac;

D = (-16) ^ 2 – 4 * 1 * 60 = 256 – 240 = 16; √D = 4;

x = (-b ± √D) / (2a);

x1 = (16 + 4) / 2 = 20/2 = 10;

x2 = (16 – 4) / 2 = 12/2 = 6;

16 – x1 = 16 – 10 = 6;

16 – x2 = 16 – 6 = 10.

The sides of a rectangle can be 10 and 6, or 6 and 10. The largest side of a rectangle is a side with a length of 10.

Answer. 10.